Q stringlengths 4 3.96k | A stringlengths 1 3k | Result stringclasses 4
values |
|---|---|---|
Lemma 5.17 Assume \( U \) is bounded, for any \( g \in {W}^{1, p}\left( U\right) \) and open \( \Omega \subseteq U,\Omega + \tau {e}_{n} \subseteq U \) for some \( \tau > 0 \) , we have \( \mathop{\lim }\limits_{{\epsilon \rightarrow 0}}{\begin{Vmatrix}g\left( y + \epsilon {e}_{n}\right) - g\left( y\right) \end{Vmatrix... | Proof: We have \( g \in {\mathcal{L}}^{p}\left( U\right) ,{Dg} \in {\mathcal{L}}^{n}\left( {U;{\mathbb{R}}^{n}}\right) \) . From Theroem 2.13, for any \( \delta > 0 \), we can find \( f \in \) \( {C}_{c}^{\infty }\left( {\mathbb{R}}^{n}\right), h \in {C}_{c}^{\infty }\left( {{\mathbb{R}}^{n};{\mathbb{R}}^{n}}\right) \)... | Yes |
Corollary 5.20 Assume \( U \) is bounded and \( \partial U \) is Lipschitz, for each \( f \in {W}^{1, p}\left( U\right) \), there exists \( {f}_{k} \in {C}^{\infty }\left( \bar{U}\right) \) such that \( \mathop{\lim }\limits_{{k \rightarrow \infty }}{\begin{Vmatrix}{f}_{k} - f\end{Vmatrix}}_{{W}^{1, p}\left( U\right) }... | Proof: Since \( \partial U \) is compact, we can cover \( \partial U \) with finite many \( C\left( {{x}_{i},\frac{{r}_{i}}{2},\frac{{h}_{i}}{2}}\right), i = 1,\cdots, m \), let \( {\Omega }_{0} = \) \( U - \mathop{\bigcup }\limits_{{i = 1}}^{m}\overline{C\left( {{x}_{i},\frac{{r}_{i}}{4},\frac{{h}_{i}}{4}}\right) } \)... | Yes |
Corollary 5.22 Assume \( U \) is bounded and \( \partial U \) is Lipschitz, there exists \( C > 0 \) and bounded open set \( V \) with \( U \subseteq \subseteq V \). For any \( f \in {C}^{1}\left( \bar{U}\right) \), there is \( g \in {W}^{1, p}\left( V\right) \) and \( \operatorname{spt}\left( g\right) \subseteq \subse... | Proof: Since \( \partial U \) is compact, we can cover \( \partial U \) with finite many \( C\left( {{x}_{i},\frac{{r}_{i}}{2},\frac{{h}_{i}}{2}}\right), i = 1,\cdots, m \), let \( {\Omega }_{0} = \) \( U - \mathop{\bigcup }\limits_{{i = 1}}^{m}\overline{C\left( {{x}_{i},\frac{{r}_{i}}{4},\frac{{h}_{i}}{4}}\right) } \)... | Yes |
Corollary 5.23 Assume \( U \) is bounded and \( \partial U \) is Lipschitz, there exists \( C > 0 \) and bounded open set \( V \) with \( U \subseteq \subseteq V \). For any \( f \in {C}^{1}\left( \bar{U}\right) \), there are \( {\left\{ {f}_{k}\right\} }_{k = 1}^{\infty } \subseteq {C}_{c}^{\infty }\left( V\right) \) ... | Proof: From Corollary 5.22, for \( f \in {C}^{1}\left( \bar{U}\right) \), we can find \( g \in {W}^{1, p}\left( V\right) \) with \( \operatorname{spt}\left( g\right) \subseteq \subseteq V \), and\n\n\[ {\left. g\right| }_{U} = {\left. f\right| }_{U},\;\parallel g{\parallel }_{{W}^{1, p}\left( V\right) } \leq C\parallel... | Yes |
Corollary 5.24 Assume \( U \) is bounded and \( \partial U \) is Lipschitz, there exists \( C > 0 \) and bounded open set \( V \) with \( U \subseteq \subseteq V \) . For any \( f \in {W}^{1, p}\left( U\right) \), there are \( {\left\{ {f}_{k}\right\} }_{k = 1}^{\infty } \subseteq {C}_{c}^{\infty }\left( V\right) \) su... | Proof: From Corollary 5.20, we can find \( {\varphi }_{k} \in {C}^{\infty }\left( \bar{U}\right) \) such that \( \mathop{\lim }\limits_{{k \rightarrow \infty }}{\begin{Vmatrix}f - {\varphi }_{k}\end{Vmatrix}}_{{W}^{1, p}\left( U\right) } = 0 \) . From Corollary 5.23, there is \( {\psi }_{kj} \in {C}_{c}^{\infty }\left(... | Yes |
Theorem 5.25 Assume \( U \subseteq {\mathbb{R}}^{n} \) is bounded set and \( \partial U \) is Lipschitz,\n\n(a) . If \( 1 \leq p < n,{\left\{ {f}_{k}\right\} }_{k = 1}^{\infty } \subseteq {W}^{1, p}\left( U\right) \) satisfies \( \mathop{\sup }\limits_{k}{\begin{Vmatrix}{f}_{k}\end{Vmatrix}}_{{W}^{1, p}\left( U\right) ... | Proof: Step (1). For any \( {f}_{j} \in {W}^{1, p}\left( U\right) \), from Corollary 5.24, we can find \( {g}_{j} \in {C}_{c}^{\infty }\left( V\right) \) such that\n\n\[ \n{\begin{Vmatrix}{f}_{k} - {g}_{k}\end{Vmatrix}}_{{W}^{1, p}\left( U\right) } \leq {2}^{-k}\;\text{ and }\;\mathop{\sup }\limits_{k}{\begin{Vmatrix}{... | Yes |
Lemma 1. Suppose \( \mathcal{F} \) is a nonempty collection of subsets of a set \( X \) such that the union of every subchain of \( \mathcal{F} \) belongs to \( \mathcal{F} \) . Suppose \( g \) is a function which associates to each \( A \in \mathcal{F} \) a set \( g\left( A\right) \in \mathcal{F} \) such that \( A \su... | Proof. Let \( {A}_{0} \in \mathcal{F} \) . Call a subcollection \( {\mathcal{F}}^{\prime } \subset \mathcal{F} \) a tower if \( {A}_{0} \in {\mathcal{F}}^{\prime } \), the union of every subchain of \( {\mathcal{F}}^{\prime } \) is in \( {\mathcal{F}}^{\prime } \), and \( g\left( A\right) \in {\mathcal{F}}^{\prime } \)... | Yes |
Lemma 2. For each \( s \in \left( {0,1}\right) \) there exists a constant \( {k}_{s} \) such that\n\n\[ \n{P}_{r}\left( {\theta - \phi }\right) \leq {k}_{s}{P}_{r}\left( {-\phi }\right) \n\]\n\nfor all \( \left( {r,\theta }\right) \) such that \( r{e}^{i\theta } \in {\Gamma }_{s} \) . | Proof. By elementary arithmetic,\n\n\[ \n{P}_{r}\left( {\theta - \phi }\right) = \frac{1 - {r}^{2}}{{\left| {e}^{i\phi } - r{e}^{i\theta }\right| }^{2}} \n\]\n\nand\n\n\[ \n{P}_{r}\left( {-\phi }\right) = \frac{1 - {r}^{2}}{{\left| {e}^{i\phi } - r\right| }^{2}}. \n\]\n\n(This alternate formula can be found in any comp... | Yes |
Lemma 3. Let \( {\left\{ {b}_{n}\right\} }_{n = 0}^{\infty } \) be a sequence of complex numbers. Then\n\n\[ \sqrt{\mathop{\sum }\limits_{{n = 0}}^{\infty }{\left| {b}_{n}\right| }^{2}} = \mathop{\sup }\limits_{{\sum {\left| {a}_{n}\right| }^{2} \leq 1}}\left| {\mathop{\sum }\limits_{{n = 0}}^{\infty }{a}_{n}{b}_{n}}\r... | Proof. If \( \sum {\left| {b}_{n}\right| }^{2} < \infty \) this follows from the Cauchy-Schwarz inequality applied to \( {\ell }^{2}\left( \mathbb{N}\right) \), where equality is achieved when \( \left\{ {a}_{n}\right\} \) is the unit vector in the same direction (actually, the conjugate) as \( \left\{ {b}_{n}\right\} ... | Yes |
Lemma 4. If \( {\mu }_{F} \) is the Borel measure corresponding to the increasing, right-continuous function \( F \), then for any \( \mu \) -measurable set \( E \) , | \[ \mu \left( E\right) = \mathop{\inf }\limits_{{E \subset \cup \left( {{a}_{j},{b}_{j}}\right) }}\sum \mu \left( \left( {{a}_{j},{b}_{j}}\right) \right) . \] In words, this lemma says that it is equivalent to use coverings of open intervals instead of half-open intervals. This is nice because it enables us to use theo... | No |
If \( {\dim }_{\mathrm{H}}F = {\overline{\dim }}_{\mathrm{B}}F \) then\n\n\[ {\dim }_{\mathrm{H}}\left( {E \times F}\right) = {\dim }_{\mathrm{H}}E + {\dim }_{\mathrm{H}}F. \] | Proof. Combining Product formulae 7.2 and 7.3 gives\n\n\[ {\dim }_{\mathrm{H}}E + {\dim }_{\mathrm{H}}F \leq {\dim }_{\mathrm{H}}\left( {E \times F}\right) \leq {\dim }_{\mathrm{H}}E + {\overline{\dim }}_{\mathrm{B}}F. \]\n\n(7.8)\n\nIt is worth noting that the basic product inequality for upper box dimensions is oppos... | No |
Proposition 14.6\n\n\\[ \nJ\\left( f\\right) = \\left\\{ {z \\in \\mathbb{C}}\\right. \\text{: the family}\\left\\{ {f}^{k}\\right\\} \\text{is not normal at}\\left. z\\right\\} \\text{.} \n\\] | Proof. If \\( z \\in J \\), then in every neighbourhood \\( V \\) of \\( z \\), there are points \\( w \\) such that \\( {f}^{k}\\left( w\\right) \\rightarrow \\infty \\), whilst \\( {f}^{k}\\left( z\\right) \\) remains bounded. Thus, no subsequence of \\( \\left\\{ {f}^{k}\\right\\} \\) is uniformly convergent on \\( ... | Yes |
Lemma 17.6\n\nFor all \( \epsilon > 0 \) ,\n\n\[ \Phi \left( {q + \delta ,\beta \left( q\right) + \left( {-\alpha + \epsilon }\right) \delta }\right) < 1 \]\n\n(17.42)\n\nand\n\n\[ \Phi \left( {q - \delta ,\beta \left( q\right) + \left( {\alpha + \epsilon }\right) \delta }\right) < 1 \]\n\n(17.43)\n\nfor all sufficient... | Proof. Recalling that \( \mathrm{d}\beta /\mathrm{d}q = - \alpha \), expansion about \( q \) gives\n\n\[ \beta \left( {q + \delta }\right) = \beta \left( q\right) - {\alpha \delta } + O\left( {\delta }^{2}\right) < \beta \left( q\right) + \left( {-\alpha + \epsilon }\right) \delta \]\nif \( \delta \) is small enough. S... | Yes |
Example 7 Let \( M\left( {X;Y}\right) \) be the set of mappings of the set \( X \) into the set \( Y \) and \( {x}_{0} \) a fixed element of \( X \) . To any function \( f \in M\left( {X;Y}\right) \) we assign its value \( f\left( {x}_{0}\right) \in Y \) at the element \( {x}_{0} \) . This relation defines a function \... | In particular, if \( Y = \mathbb{R} \), that is, \( Y \) is the set of real numbers, then to each function \( f : X \rightarrow \mathbb{R} \) the function \( F : M\left( {X;\mathbb{R}}\right) \rightarrow \mathbb{R} \) assigns the number \( F\left( f\right) = f\left( {x}_{0}\right) \) . Thus \( F \) is a function define... | Yes |
The position of a particle in space is determined by an ordered triple of numbers \( \left( {x, y, z}\right) \) called its spatial coordinates. The set of all such ordered triples can be thought of as the direct product \( \mathbb{R} \times \mathbb{R} \times \mathbb{R} = {\mathbb{R}}^{3} \) of three real lines \( \math... | A particle in motion is located at some point of the space \( {\mathbb{R}}^{3} \) having coordinates \( \left( {x\left( t\right), y\left( t\right), z\left( t\right) }\right) \) at each instant \( t \) of time. Thus the motion of a particle can be interpreted as a mapping \( \gamma : \mathbb{R} \rightarrow {\mathbb{R}}^... | Yes |
The kinetic energy \( K \) of a system of \( n \) material particles depends on their velocities. The total mechanical energy of the system \( E \), defined as \( E = K + U \) , that is, the sum of the kinetic and potential energies, thus depends on both the configuration \( q \) of the system and the set of velocities... | The total mechanical energy of the system is therefore a function \( E : \Phi \rightarrow \mathbb{R} \) defined on the subset \( \Phi \) of the phase space \( {\mathbb{R}}^{6n} \) and assuming values in the domain \( \mathbb{R} \) of real numbers.\n\nIn particular, if the system is isolated, that is, no external forces... | Yes |
Example 13 The diagonal\n\n\[ \Delta = \left\{ {\left( {a, b}\right) \in {X}^{2} \mid a = b}\right\} \] | is a subset of \( {X}^{2} \) defining the relation of equality between elements of \( X \) . Indeed, \( {a\Delta b} \) means that \( \left( {a, b}\right) \in \Delta \), that is, \( a = b \) . | Yes |
Example 14 Let \( X \) be the set of lines in a plane. Two lines \( a \in X \) and \( b \in X \) will be considered to be in the relation \( \mathcal{R} \), and we shall write \( a\mathcal{R}b \), if \( b \) is parallel to \( a \). It is clear that this condition distinguishes a set \( \mathcal{R} \) of pairs \( \left(... | \( a\mathcal{R}a \) (reflexivity);\n\n\( a\mathcal{R}b \Rightarrow b\mathcal{R}a \) (symmetry);\n\n\( \left( {a\mathcal{R}b}\right) \land \left( {b\mathcal{R}c}\right) \Rightarrow a\mathcal{R}c \) (transitivity). | Yes |
Example 15 Let \( M \) be a set and \( X = \mathcal{P}\left( M\right) \) the set of its subsets. For two arbitrary elements \( a \) and \( b \) of \( X = \mathcal{P}\left( M\right) \), that is, for two subsets \( a \) and \( b \) of \( M \), one of the following three possibilities always holds: \( a \) is contained in... | As an example of a relation \( \mathcal{R} \) on \( {X}^{2} \), consider the relation of inclusion for subsets of \( M \), that is, make the definition\n\n\[ \n a\mathcal{R}b \mathrel{\text{:=}} \left( {a \subset b}\right) .\n\]\n\nThis relation obviously has the following properties:\n\n\( a\mathcal{R}a \) (reflexivit... | Yes |
Example 5 \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{1}{{q}^{n}} = 0 \) if \( \left| q\right| > 1 \) . | Let us verify this last assertion using the definition of the limit. As was shown in Paragraph c of Sect. 2.2.4, for every \( \varepsilon > 0 \) there exists \( N \in \mathbb{N} \) such that \( \frac{1}{{\left| q\right| }^{N}} < \varepsilon \) . Since \( \left| q\right| > 1 \), we shall have \( \left| {\frac{1}{{q}^{n}... | Yes |
The sequence \( 1,2,\frac{1}{3},4,\frac{1}{5},6,\frac{1}{7},\ldots \) whose \( n \) th term is \( {x}_{n} = {n}^{{\left( -1\right) }^{n}} \) , \( n \in \mathbb{N} \), is divergent. | Indeed, if \( A \) were the limit of this sequence, then, as follows from the definition of limit, any neighborhood of \( A \) would contain all but a finite number of terms of the sequence.\n\nA number \( A \neq 0 \) cannot be the limit of this sequence; for if \( \varepsilon = \frac{\left| A\right| }{2} > 0 \), all t... | Yes |
Theorem 4 (Cauchy’s convergence criterion) A numerical sequence converges if and only if it is a Cauchy sequence. | Proof Suppose \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = A \) . Given \( \varepsilon > 0 \), we find an index \( N \) such that \( \mid {x}_{n} - \) \( \left| A\right| < \frac{\varepsilon }{2} \) for \( n > N \) . Then if \( m > N \) and \( n > N \), we have \( \left| {{x}_{m} - {x}_{n}}\right| \leq \l... | Yes |
Example 8 The sequence \( {\left( -1\right) }^{n}\left( {n = 1,2,\ldots }\right) \) has no limit, since it is not a Cauchy sequence. | Even though this fact is obvious, we shall give a formal verification. The negation of the statement that \( \left\{ {x}_{n}\right\} \) is a Cauchy sequence is the following:\n\n\[\n\exists \varepsilon > 0\forall N \in \mathbb{N}\exists n > N\exists m > N\left( {\left| {{x}_{m} - {x}_{n}}\right| \geq \varepsilon }\righ... | Yes |
Example 9 Let\n\n\\[ \n{x}_{1} = 0.{\alpha }_{1},\;{x}_{2} = 0.{\alpha }_{1}{\alpha }_{2},\;{x}_{3} = 0.{\alpha }_{1}{\alpha }_{2}{\alpha }_{3},\;\ldots ,\;{x}_{n} = 0.{\alpha }_{1}{\alpha }_{2}\ldots {\alpha }_{n},\;\ldots \n\\]\n\nbe a sequence of finite binary fractions in which each successive fraction is obtained ... | Let \( m > n \) . Let us estimate the difference \( {x}_{m} - {x}_{n} \) :\n\n\\[ \n\left| {{x}_{m} - {x}_{n}}\right| = \left| {\frac{{\alpha }_{n + 1}}{{2}^{n + 1}} + \cdots + \frac{{\alpha }_{m}}{{2}^{m}}}\right| \leq \n\\]\n\n\\[ \n\leq \frac{1}{{2}^{n + 1}} + \cdots + \frac{1}{{2}^{m}} = \frac{{\left( \frac{1}{2}\r... | Yes |
Example 10 Consider the sequence \( \left\{ {x}_{n}\right\} \), where\n\n\[ \n{x}_{n} = 1 + \frac{1}{2} + \cdots + \frac{1}{n} \n\] | Since\n\n\[ \n\left| {{x}_{2n} - {x}_{n}}\right| = \frac{1}{n + 1} + \cdots + \frac{1}{n + n} > n \cdot \frac{1}{2n} = \frac{1}{2}, \n\]\n\nfor all \( n \in \mathbb{N} \), the Cauchy criterion implies immediately that this sequence does not have a limit. | Yes |
Theorem 5 (Weierstrass) In order for a nondecreasing sequence to have a limit it is necessary and sufficient that it be bounded above. | Proof The fact that any convergent sequence is bounded was proved above under general properties of the limit of a sequence. For that reason only the sufficiency assertion is of interest.\n\nBy hypothesis the set of values of the sequence \( \left\{ {x}_{n}\right\} \) is bounded above and hence has a least upper bound ... | Yes |
Example 11 \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{n}{{q}^{n}} = 0 \) if \( q > 1 \) . | Proof Indeed, if \( {x}_{n} = \frac{n}{{q}^{n}} \), then \( {x}_{n + 1} = \frac{n + 1}{nq}{x}_{n} \) for \( n \in \mathbb{N} \) . Since \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{n + 1}{nq} = \) \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {1 + \frac{1}{n}}\right) \frac{1}{q} = \mathop{\lim }\l... | Yes |
\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\sqrt[n]{n} = 1 \] | Proof By what was just proved, for a given \( \varepsilon > 0 \) there exists \( N \in \mathbb{N} \) such that \( 1 \leq n < {\left( 1 + \varepsilon \right) }^{n} \) for all \( n > N \) . Then for \( n > N \) we obtain \( 1 \leq \sqrt[n]{n} < 1 + \varepsilon \) and hence \( \mathop{\lim }\limits_{{n \rightarrow \infty ... | No |
\[ \mathop{\lim }\limits_{{n \rightarrow \infty }}\sqrt[n]{a} = 1\;\text{ for any }a > 0. \] | Proof Assume first that \( a \geq 1 \) . For any \( \varepsilon > 0 \) there exists \( N \in \mathbb{N} \) such that \( 1 \leq a < \) \( {\left( 1 + \varepsilon \right) }^{n} \) for all \( n > N \), and we then have \( 1 \leq \sqrt[n]{a} < 1 + \varepsilon \) for all \( n > N \), which says \( \mathop{\lim }\limits_{{n ... | Yes |
Example 12 \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{{q}^{n}}{n!} = 0 \) ; here \( q \) is any real number, \( n \in \mathbb{N} \), and \( n! \mathrel{\text{:=}} 1 \cdot 2 \) . \( \ldots \cdot n \) . | Proof If \( q = 0 \), the assertion is obvious. Further, since \( \left| \frac{{q}^{n}}{n!}\right| = \frac{{\left| q\right| }^{n}}{n!} \), it suffices to prove the assertion for \( q > 0 \) . Reasoning as in Example 11, we remark that \( {x}_{n + 1} = \) \( \frac{q}{n + 1}{x}_{n} \) . Since the set of natural numbers i... | Yes |
Lemma 1 (Bolzano-Weierstrass) Every bounded sequence of real numbers contains a convergent subsequence. | Proof Let \( E \) be the set of values of the bounded sequence \( \left\{ {x}_{n}\right\} \) . If \( E \) is finite, there exists a point \( x \in E \) and a sequence \( {n}_{1} < {n}_{2} < \cdots \) of indices such that \( {x}_{{n}_{1}} = {x}_{{n}_{2}} = \) \( \cdots = x \) . The subsequence \( \left\{ {x}_{{n}_{k}}\r... | Yes |
Lemma 2 From each sequence of real numbers one can extract either a convergent subsequence or a subsequence that tends to infinity. | Proof The new case here occurs when the sequence \( \left\{ {x}_{n}\right\} \) is not bounded. Then for each \( k \in \mathbb{N} \) we can choose \( {n}_{k} \in \mathbb{N} \) such that \( \left| {x}_{{n}_{k}}\right| > k \) and \( {n}_{k} < {n}_{k + 1} \) . We then obtain a subsequence \( \left\{ {x}_{{n}_{k}}\right\} \... | Yes |
Example 14 \( {x}_{k} = {\left( -1\right) }^{k}, k \in \mathbb{N} \) : | \[ \mathop{\lim }\limits_{{k \rightarrow \infty }}{x}_{k} = \mathop{\lim }\limits_{{n \rightarrow \infty }}\mathop{\inf }\limits_{{k \geq n}}{x}_{k} = \mathop{\lim }\limits_{{n \rightarrow \infty }}\mathop{\inf }\limits_{{k \geq n}}{\left( -1\right) }^{k} = \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {-1}\rig... | Yes |
Example 15 \( {x}_{k} = {k}^{{\left( -1\right) }^{k}}, k \in \mathbb{N} \) : | \[ \mathop{\lim }\limits_{{k \rightarrow \infty }}{k}^{{\left( -1\right) }^{k}} = \mathop{\lim }\limits_{{n \rightarrow \infty }}\mathop{\inf }\limits_{{k \geq n}}{k}^{{\left( -1\right) }^{k}} = \mathop{\lim }\limits_{{n \rightarrow \infty }}0 = 0, \] \[ \mathop{\lim }\limits_{{k \rightarrow \infty }}{k}^{{\left( -1\ri... | Yes |
Example 16 \( {x}_{k} = k, k \in \mathbb{N} \) : | \[ \mathop{\lim }\limits_{{k \rightarrow \infty }}k = \mathop{\lim }\limits_{{n \rightarrow \infty }}\mathop{\inf }\limits_{{k \geq n}}k = \mathop{\lim }\limits_{{n \rightarrow \infty }}n = + \infty , \] \[ \mathop{\lim }\limits_{{k \rightarrow \infty }}k = \mathop{\lim }\limits_{{n \rightarrow \infty }}\mathop{\sup }\... | Yes |
Example 17 \( {x}_{k} = \frac{{\left( -1\right) }^{k}}{k}, k \in \mathbb{N} \) | \[ \mathop{\lim }\limits_{{k \rightarrow \infty }}\frac{{\left( -1\right) }^{k}}{k} = \mathop{\lim }\limits_{{n \rightarrow \infty }}\mathop{\inf }\limits_{{k \geq n}}\frac{{\left( -1\right) }^{k}}{k} = \mathop{\lim }\limits_{{n \rightarrow \infty }}\left\{ \begin{array}{ll} - \frac{1}{n}, & \text{ if }n = {2m} + 1 \\ ... | Yes |
Example 18 \( {x}_{k} = - {k}^{2}, k \in \mathbb{N} \) : | \[ \mathop{\lim }\limits_{{k \rightarrow \infty }}\left( {-{k}^{2}}\right) = \mathop{\lim }\limits_{{n \rightarrow \infty }}\mathop{\inf }\limits_{{k \geq n}}\left( {-{k}^{2}}\right) = - \infty . \] | Yes |
Example 19 \( {x}_{k} = {\left( -1\right) }^{k}k, k \in \mathbb{N} \) : | \[ \mathop{\lim }\limits_{{k \rightarrow \infty }}{\left( -1\right) }^{k}k = \mathop{\lim }\limits_{{n \rightarrow \infty }}\mathop{\inf }\limits_{{k \geq n}}{\left( -1\right) }^{k}k = \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {-\infty }\right) = - \infty , \] \[ \overline{\mathop{\lim }\limits_{{k \rightar... | Yes |
Proposition 1 The inferior and superior limits of a bounded sequence are respectively the smallest and largest partial limits of the sequence. | Proof Let us prove this, for example, for the inferior limit \( i = \mathop{\lim }\limits_{{k \rightarrow \infty }}{x}_{k} \) . What we know about the sequence \( {i}_{n} = \mathop{\inf }\limits_{{k \geq n}}{x}_{k} \) is that it is nondecreasing and that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{i}_{n} = i \in... | Yes |
Corollary 3 A sequence has a limit or tends to negative or positive infinity if and only if its inferior and superior limits are the same. | Proof The cases when \( \mathop{\lim }\limits_{{k \rightarrow \infty }}{x}_{k} = {\overline{\lim }}_{k \rightarrow \infty }{x}_{k} = + \infty \) or \( \mathop{\lim }\limits_{{k \rightarrow \infty }}{x}_{k} = \) \( {\overline{\lim }}_{k \rightarrow \infty }{x}_{k} = - \infty \) have been investigated above, and so we ma... | Yes |
Corollary 4 A sequence converges if and only if every subsequence of it converges. | Proof The inferior and superior limits of a subsequence lie between those of the sequence itself. If the sequence converges, its inferior and superior limits are the same, and so those of the subsequence must also be the same, proving that the subsequence converges. Moreover, the limit of the subsequence must be the sa... | Yes |
Corollary 5 The Bolzano-Weierstrass Lemma in its restricted and wider formulations follows from Propositions 1 and \( {1}^{\prime } \) respectively. | Proof Indeed, if the sequence \( \left\{ {x}_{k}\right\} \) is bounded, then the points \( i = \mathop{\lim }\limits_{{k \rightarrow \infty }}{x}_{k} \) and \( s = {\overline{\lim }}_{k \rightarrow \infty }{x}_{k} \) are finite and, by what has been proved, are partial limits of the sequence. Only when \( i = s \) does... | No |
Corollary 6 If only a finite number of terms of a series are changed, the resulting new series will converge if the original series did and diverge if it diverged. | Proof For the proof it suffices to assume that the number \( N \) in the Cauchy convergence criterion is larger than the largest index among the terms that were altered. \( ▱ \) | No |
Corollary 7 A necessary condition for convergence of the series \( {a}_{1} + \cdots + {a}_{n} + \cdots \) is that the terms tend to zero as \( n \rightarrow \infty \), that is, it is necessary that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{a}_{n} = 0 \) . | Proof It suffices to set \( m = n \) in the Cauchy convergence criterion and use the definition of the limit of a sequence.\n\nHere is another proof: \( {a}_{n} = {s}_{n} - {s}_{n - 1} \), and, given that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{s}_{n} = s \), we have \( \mathop{\lim }\limits_{{n \rightarrow ... | Yes |
The series \( 1 + q + {q}^{2} + \cdots + {q}^{n} + \cdots \) is often called the geometric series. Let us investigate its convergence. | Since \( \left| {q}^{n}\right| = {\left| q\right| }^{n} \), we have \( \left| {q}^{n}\right| \geq 1 \) when \( \left| q\right| \geq 1 \), and in this case the necessary condition for convergence is not met.\n\nNow suppose \( \left| q\right| < 1 \) . Then\n\n\[ \n{s}_{n} = 1 + q + \cdots + {q}^{n - 1} = \frac{1 - {q}^{n... | Yes |
The series \( 1 + \frac{1}{2} + \cdots + \frac{1}{n} + \cdots \) is called the harmonic series, since each term from the second on is the harmonic mean of the two terms on either side of it (see Exercise 6 at the end of this section). | The terms of the series tend to zero, but the sequence of partial sums\n\n\[ \n{s}_{n} = 1 + \frac{1}{2} + \cdots + \frac{1}{n} \n\]\n\nas was shown in Example 10, diverges. This means that in this case \( {s}_{n} \rightarrow + \infty \) as \( n \rightarrow \infty \).\n\nThus the harmonic series diverges. | No |
The series \( 1 - 1 + \frac{1}{2} - \frac{1}{2} + \frac{1}{3} - \frac{1}{3} + \cdots \), whose partial sums are either \( \frac{1}{n} \) or 0, converges to 0. | At the same time, the series of absolute values of its terms\n\n\[ 1 + 1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{3} + \frac{1}{3} + \cdots \]\n\ndiverges, as follows from the Cauchy convergence criterion, just as in the case of the harmonic series:\n\n\[ \left| {\frac{1}{n + 1} + \frac{1}{n + 1} + \cdots + \frac{1}{n + ... | No |
Theorem 7 (Criterion for convergence of series of nonnegative terms) A series \( {a}_{1} + \) \( \cdots + {a}_{n} + \cdots \) whose terms are nonnegative converges if and only if the sequence of partial sums is bounded above. | Proof This follows from the definition of convergence of a series and the criterion for convergence of a nondecreasing sequence, which the sequence of partial sums is, in this case: \( {s}_{1} \leq {s}_{2} \leq \cdots \leq {s}_{n} \leq \cdots \) . | Yes |
Theorem 8 (Comparison theorem) Let \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n} \) and \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{b}_{n} \) be two series with nonnegative terms. If there exists an index \( N \in \mathbb{N} \) such that \( {a}_{n} \leq {b}_{n} \) for all \( n > N \) , then the convergence of th... | Proof Since a finite number of terms has no effect on the convergence of a series, we can assume with no loss of generality that \( {a}_{n} \leq {b}_{n} \) for every index \( n \in \mathbb{N} \) . Then \( {A}_{n} = \mathop{\sum }\limits_{{k = 1}}^{n}{a}_{k} \leq \mathop{\sum }\limits_{{k = 1}}^{n}{b}_{k} = {B}_{n} \) .... | Yes |
Since \( \frac{1}{n\left( {n + 1}\right) } < \frac{1}{{n}^{2}} < \frac{1}{\left( {n - 1}\right) n} \) for \( n \geq 2 \), we conclude that the series \( \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{n}^{2}} \) and \( \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{n\left( {n + 1}\right) } \) converge or diverge ... | But the latter series can be summed directly, by observing that \( \frac{1}{k\left( {k + 1}\right) } = \frac{1}{k} - \frac{1}{k + 1} \) , and therefore \( \mathop{\sum }\limits_{{k = 1}}^{n}\frac{1}{k\left( {k + 1}\right) } = 1 - \frac{1}{n + 1} \) . Hence \( \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{n\left( {n... | Yes |
Example 25 It should be observed that the comparison theorem applies only to series with nonnegative terms. | Indeed, if we set \( {a}_{n} = - n \) and \( {b}_{n} = 0 \), for example, we have \( {a}_{n} < {b}_{n} \) and the series \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{b}_{n} \) converges while \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n} \) diverges. | Yes |
Corollary 8 (The Weierstrass \( M \) -test for absolute convergence) Let \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n} \) and \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{b}_{n} \) be series. Suppose there exists an index \( N \in \mathbb{N} \) such that \( \left| {a}_{n}\right| \leq {b}_{n} \) for all \( n > N \... | Proof In fact, by the comparison theorem the series \( \mathop{\sum }\limits_{{n = 1}}^{\infty }\left| {a}_{n}\right| \) will then converge, and that is what is meant by the absolute convergence of \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n} \) . | Yes |
The series \( \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{\sin n}{{n}^{2}} \) converges absolutely. | since \( \left| \frac{\sin n}{{n}^{2}}\right| \leq \frac{1}{{n}^{2}} \) and the series \( \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{{n}^{2}} \) converges, as we saw in Example 24. | Yes |
Corollary 9 (Cauchy’s test) Let \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n} \) be a given series and \( \alpha = {\overline{\lim }}_{n \rightarrow \infty }\sqrt[n]{\left| {a}_{n}\right| } \) . Then the following are true:\n\na) if \( \alpha < 1 \), the series \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n} ... | Proof a) If \( \alpha < 1 \), we can choose \( q \in \mathbb{R} \) such that \( \alpha < q < 1 \) . Fixing \( q \), by definition of the superior limit, we find \( N \in \mathbb{N} \) such that \( \sqrt[n]{\left| {a}_{n}\right| } < q \) for all \( n > N \) . Thus we shall have \( \left| {a}_{n}\right| < {q}^{n} \) for ... | Yes |
Let us investigate the values of \( x \in \mathbb{R} \) for which the series\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{\infty }{\left( 2 + {\left( -1\right) }^{n}\right) }^{n}{x}^{n} \]\n\nconverges. | We compute \( \alpha = {\overline{\lim }}_{n \rightarrow \infty }\sqrt[n]{\left| {\left( 2 + {\left( -1\right) }^{n}\right) }^{n}{x}^{n}\right| } = \left| x\right| {\overline{\lim }}_{n \rightarrow \infty }\left| {2 + {\left( -1\right) }^{n}}\right| = 3\left| x\right| \) . Thus for \( \left| x\right| < \frac{1}{3} \) t... | Yes |
Corollary 10 (d’Alembert’s \( {}^{6} \) test) Suppose the limit \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left| \frac{{a}_{n + 1}}{{a}_{n}}\right| = \alpha \) exists for the series \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n} \) . Then,\n\na) if \( \alpha < 1 \), the series \( \mathop{\sum }\limits_{{n ... | Proof a) If \( \alpha < 1 \), there exists a number \( q \) such that \( \alpha < q < 1 \) . Fixing \( q \) and using properties of limits, we find an index \( N \in \mathbb{N} \) such that \( \left| \frac{{a}_{n + 1}}{{a}_{n}}\right| < q \) for \( n > N \) . Since a finite number of terms has no effect on the converge... | Yes |
Let us determine the values of \( x \in \mathbb{R} \) for which the series\n\n\[ \mathop{\sum }\limits_{{n = 1}}^{\infty }\frac{1}{n!}{x}^{n} \]\n\nconverges. | For \( x = 0 \) it obviously converges absolutely.\n\nFor \( x \neq 0 \) we have \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left| \frac{{a}_{n + 1}}{{a}_{n}}\right| = \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{\left| x\right| }{n + 1} = 0 \) .\n\nThus, this series converges absolutely for every value ... | Yes |
Proposition 2 (Cauchy) If \( {a}_{1} \geq {a}_{2} \geq \cdots \geq 0 \), the series \( \mathop{\sum }\limits_{{n = 1}}^{\infty }{a}_{n} \) converges if and only if the series \( \mathop{\sum }\limits_{{k = 0}}^{\infty }{2}^{k}{a}_{{2}^{k}} = {a}_{1} + 2{a}_{2} + 4{a}_{4} + 8{a}_{8} + \cdots \) converges. | Proof Since\n\n\[ \n{a}_{2} \leq {a}_{2} \leq {a}_{1} \n\]\n\n\[ \n2{a}_{4} \leq {a}_{3} + {a}_{4} \leq 2{a}_{2} \n\]\n\n\[ \n4{a}_{8} \leq {a}_{5} + {a}_{6} + {a}_{7} + {a}_{8} \leq 4{a}_{4} \n\]\n\n\[ \n{2}^{n}{a}_{{2}^{n + 1}} \leq {a}_{{2}^{n} + 1} + \cdots + {a}_{{2}^{n + 1}} \leq {2}^{n}{a}_{{2}^{n}} \n\]\n\nby a... | Yes |
Let \( E = \mathbb{R} \smallsetminus 0 \), and \( f\left( x\right) = x\sin \frac{1}{x} \). We shall verify that\n\n\[\n\mathop{\lim }\limits_{{E \ni x \rightarrow 0}}x\sin \frac{1}{x} = 0\n\] | Indeed, for a given \( \varepsilon > 0 \) we choose \( \delta = \varepsilon \). Then for \( 0 < \left| x\right| < \delta = \varepsilon \), taking account of the inequality \( \left| {x\sin \frac{1}{x}}\right| \leq \left| x\right| \), we shall have \( \left| {x\sin \frac{1}{x}}\right| < \varepsilon \). | Yes |
Example 3 Let us show that \( \mathop{\lim }\limits_{{x \rightarrow 0}}\left| {\operatorname{sgn}x}\right| = 1 \) . | Indeed, for \( x \in \mathbb{R} \smallsetminus 0 \) we have \( \left| {\operatorname{sgn}x}\right| = 1 \), that is, the function is constant and equal to 1 in any deleted neighborhood \( \mathring{U}\left( 0\right) \) of 0 . Hence for any neighborhood \( V\left( 1\right) \) we obtain \( f\left( {\overset{ \circ }{U}\le... | Yes |
Example 4 We saw in Example 2 that the limit \( \mathop{\lim }\limits_{{\mathbb{R} \ni x \rightarrow 0}}\operatorname{sgn}x \) does not exist. Remarking, however, that the restriction \( \operatorname{sgn}{|}_{\mathbb{R} - } \) of sgn to \( {\mathbb{R}}_{ - } \) is a constant function equal to -1 and \( \operatorname{s... | \[ \mathop{\lim }\limits_{{{\mathbb{R}}_{ - } \ni x \rightarrow 0}}\operatorname{sgn}x = - 1,\;\text{ and }\;\mathop{\lim }\limits_{{{\mathbb{R}}_{ + } \ni x \rightarrow 0}}\operatorname{sgn}x = 1, \] | Yes |
Developing the idea of Example 2, one can show similarly that \( \sin \frac{1}{x} \) has no limit as \( x \rightarrow 0 \) . | Indeed, in any deleted neighborhood \( \overset{ \circ }{U}\left( 0\right) \) of 0 there are always points of the form \( \frac{1}{-\pi /2 + {2\pi n}} \) and \( \frac{1}{\pi /2 + {2\pi n}} \), where \( n \in \mathbb{N} \). At these points the function assumes the values -1 and 1 respectively. But these two numbers cann... | Yes |
Proposition 1 9 The relation \( \mathop{\lim }\limits_{{E \ni x \rightarrow a}}f\left( x\right) = A \) holds if and only if for every sequence \( \left\{ {x}_{n}\right\} \) of points \( {x}_{n} \in E \smallsetminus a \) converging to \( a \), the sequence \( \left\{ {f\left( {x}_{n}\right) }\right\} \) converges to \( ... | Proof The fact that \( \left( {\mathop{\lim }\limits_{{E \ni x \rightarrow a}}f\left( x\right) = A}\right) \Rightarrow \left( {\mathop{\lim }\limits_{{n \rightarrow \infty }}f\left( {x}_{n}\right) = A}\right) \) follows immediately from the definitions. Indeed, if \( \mathop{\lim }\limits_{{E \ni x \rightarrow a}}f\lef... | Yes |
c) \( \left( {\mathop{\lim }\limits_{{E \ni x \rightarrow a}}f\left( x\right) = {A}_{1}}\right) \land \left( {\mathop{\lim }\limits_{{E \ni x \rightarrow a}}f\left( x\right) = {A}_{2}}\right) \Rightarrow \left( {{A}_{1} = {A}_{2}}\right) \) | Proof The assertion a) that an ultimately constant function has a limit, and assertion b) that a function having a limit is ultimately bounded, follow immediately from the corresponding definitions. We now turn to the proof of the uniqueness of the limit.\n\nSuppose \( {A}_{1} \neq {A}_{2} \) . Choose neighborhoods \( ... | Yes |
Theorem 2 Let \( f : E \rightarrow \mathbb{R} \) and \( g : E \rightarrow \mathbb{R} \) be two functions with a common domain of definition.\n\nIf \( \mathop{\lim }\limits_{{E \ni x \rightarrow a}}f\left( x\right) = A \) and \( \mathop{\lim }\limits_{{E \ni x \rightarrow a}}g\left( x\right) = B \), then\n\n a) \( \math... | As already noted at the beginning of Sect. 3.2.2, this theorem is an immediate consequence of the corresponding theorem on limits of sequences, given Proposition 1. The theorem can also be obtained by repeating the proof of the theorem on the algebraic properties of the limit of a sequence. The changes needed in the pr... | No |
\[ \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\sin x}{x} = 1 \] | Proof Assuming that \( \left| x\right| < \pi /2 \), from the inequality in a) we have\n\n\[ 1 - {\sin }^{2}x < \frac{\sin x}{x} < 1 \]\n\nBut \( \mathop{\lim }\limits_{{x \rightarrow 0}}\left( {1 - {\sin }^{2}x}\right) = 1 - \mathop{\lim }\limits_{{x \rightarrow 0}}\sin x \cdot \mathop{\lim }\limits_{{x \rightarrow 0}}... | Yes |
Theorem 4 (The Cauchy criterion for the existence of a limit of a function) Let \( X \) be a set and \( \mathcal{B} \) a base in \( X \). A function \( f : X \rightarrow \mathbb{R} \) has a limit over the base \( \mathcal{B} \) if and only if for every \( \varepsilon > 0 \) there exits \( B \in \mathcal{B} \) such that... | Proof Necessity. If \( \mathop{\lim }\limits_{\mathcal{B}}f\left( x\right) = A \in \mathbb{R} \), then, for all \( \varepsilon > 0 \), there exists an element \( B \in \mathcal{B} \) such that \( \left| {f\left( x\right) - A}\right| < \varepsilon /3 \) for all \( x \in B \). But then, for any \( {x}_{1},{x}_{2} \in B \... | Yes |
We shall show that when \( X = \mathbb{N} \) and \( \mathcal{B} \) is the base \( n \rightarrow \infty, n \in \mathbb{N} \) , the general Cauchy criterion just proved for the existence of the limit of a function coincides with the Cauchy criterion already studied for the existence of a limit of a sequence. | Indeed, an element of the base \( n \rightarrow \infty, n \in \mathbb{N} \), is a set \( B = \mathbb{N} \cap U\left( \infty \right) = \{ n \in \mathbb{N} \mid \) \( N < n\} \) consisting of the natural numbers \( n \in \mathbb{N} \) larger than some number \( N \in \mathbb{R} \) . Without loss of generality we may assu... | Yes |
Theorem 5 (The limit of a composite function) Let \( Y \) be a set, \( {\mathcal{B}}_{Y} \) a base in \( Y \), and \( g : Y \rightarrow \mathbb{R} \) a mapping having a limit over the base \( {\mathcal{B}}_{Y} \) . Let \( X \) be a set, \( {\mathcal{B}}_{X} \) a base in \( X \) and \( f : X \rightarrow Y \) a mapping o... | Proof The composite function \( g \circ f : X \rightarrow \mathbb{R} \) is defined, since \( f\left( X\right) \subset Y \) . Suppose \( \mathop{\lim }\limits_{{\mathcal{B}}_{Y}}g\left( y\right) = A \) . We shall show that \( \mathop{\lim }\limits_{{\mathcal{B}}_{X}}\left( {g \circ f}\right) \left( x\right) = A \) . Giv... | Yes |
Let us find the following limit:\n\n\[ \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\sin {7x}}{7x} = ? \] | If we set \( g\left( y\right) = \frac{\sin y}{y} \) and \( f\left( x\right) = {7x} \), then \( \left( {g \circ f}\right) \left( x\right) = \frac{\sin {7x}}{7x} \) . In this case \( Y = \) \( \mathbb{R} \smallsetminus 0 \) and \( X = \mathbb{R} \) . Since \( \mathop{\lim }\limits_{{y \rightarrow 0}}g\left( y\right) = \m... | Yes |
The function \( \left( {g \circ f}\right) \left( x\right) = \left| {\operatorname{sgn}\left( {x\sin \frac{1}{x}}\right) }\right| \) has no limit as \( x \rightarrow 0 \) . | Indeed, in any deleted neighborhood of \( x = 0 \) there are zeros of the function \( \sin \frac{1}{x} \), so that the function \( \left| {\operatorname{sgn}\left( {x\sin \frac{1}{x}}\right) }\right| \) assumes both the value 1 and the value 0 in any such neighborhood. By the Cauchy criterion, this function cannot have... | Yes |
\[ \mathop{\lim }\limits_{{x \rightarrow \infty }}{\left( 1 + \frac{1}{x}\right) }^{x} = \mathrm{e}. \] | Proof Let us make the following assumptions:\n\n\[ Y = \mathbb{N},\;{\mathcal{B}}_{Y}\text{ is the base }n \rightarrow \infty, n \in \mathbb{N}; \]\n\n\[ X = {\mathbb{R}}_{ + } = \{ x \in \mathbb{R} \mid x > 0\} ,\;{\mathcal{B}}_{X}\text{ is the base }x \rightarrow + \infty ; \]\n\n\[ f : X \rightarrow Y\;\text{ is the... | Yes |
\[ \mathop{\lim }\limits_{{t \rightarrow 0}}{\left( 1 + t\right) }^{1/t} = \mathrm{e}. \] | Proof After the substitution \( x = 1/t \), we return to the limit considered in the preceding example. | No |
\[ \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{x}{{q}^{x}} = 0,\;\text{ if }q > 1 \] | Proof We know (see Example 11 in Sect. 3.1) that \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\frac{n}{{q}^{n}} = 0 \) if \( q > 1 \) . Now, as in Example 20, we can consider the auxiliary mapping \( f : {\mathbb{R}}_{ + } \rightarrow \mathbb{N} \) given by the function \( \left\lbrack x\right\rbrack \) (the integ... | Yes |
\[ \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{{\log }_{a}x}{x} = 0 \] | Proof Let \( a > 1 \) . Set \( t = {\log }_{a}x \), so that \( x = {a}^{t} \) . From the properties of the exponential function and the logarithm (taking account of the unboundedness of \( {a}^{n} \) for \( n \in \mathbb{N} \) ) we have \( \left( {x \rightarrow + \infty }\right) \Leftrightarrow \left( {t \rightarrow + ... | Yes |
A necessary and sufficient condition for a function \( f : E \rightarrow \mathbb{R} \) that is nondecreasing on the set \( E \) to have a limit as \( x \rightarrow s, x \in E \), is that it be bounded above. For this function to have a limit as \( x \rightarrow i, x \in E \), it is necessary and sufficient that it be b... | We shall prove this theorem for the limit \( \mathop{\lim }\limits_{{E \ni x \rightarrow s}}f\left( x\right) \). If this limit exists, then, like any function having a limit, the function \( f \) is ultimately bounded over the base \( E \ni x \rightarrow s \). Since \( f \) is nondecreasing on \( E \), it follows that ... | Yes |
Example 24 \( {x}^{2} = o\left( x\right) \) as \( x \rightarrow 0 \) | since \( {x}^{2} = x \cdot x \) | No |
We shall show that for \( a > 1 \) and any \( n \in \mathbb{Z} \)\n\n\[ \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{{x}^{n}}{{a}^{x}} = 0 \]\n\nthat is, \( {x}^{n} = o\left( {a}^{x}\right) \) as \( x \rightarrow + \infty \) . | Proof If \( n \leq 0 \) the assertion is obvious. If \( n \in \mathbb{N} \), then, setting \( q = \sqrt[n]{a} \), we have \( q > 1 \) and \( \frac{{x}^{n}}{{a}^{x}} = {\left( \frac{x}{{q}^{x}}\right) }^{n} \), and therefore\n\n\[ \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{{x}^{n}}{{a}^{x}} = \mathop{\lim }\... | Yes |
\[ \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{{x}^{\alpha }}{{a}^{x}} = 0 \] for \( a > 1 \) and any \( \alpha \in \mathbb{R} \), that is, \( {x}^{\alpha } = o\left( {a}^{x}\right) \) as \( x \rightarrow + \infty \) . | Indeed, let us choose \( n \in \mathbb{N} \) such that \( n > \alpha \) . Then for \( x > 1 \) we obtain \[ 0 < \frac{{x}^{\alpha }}{{a}^{x}} < \frac{{x}^{n}}{{a}^{x}} \] Using properties of the limit and the result of the preceding example, we find that \( \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{{x}^{\a... | Yes |
\[ \mathop{\lim }\limits_{{{\mathbb{R}}_{ + } \ni x \rightarrow 0}}\frac{{a}^{-1/x}}{{x}^{\alpha }} = 0 \] for \( a > 1 \) and any \( \alpha \in \mathbb{R} \), that is, \( {a}^{-1/x} = o\left( {x}^{\alpha }\right) \) as \( x \rightarrow 0, x \in {\mathbb{R}}_{ + } \) . | Proof Setting \( x = - 1/t \) in this case and using the theorem on the limit of a composite function and the result of the preceding example, we find \[ \mathop{\lim }\limits_{{{\mathbb{R}}_{ + } \ni x \rightarrow 0}}\frac{{a}^{-1/x}}{{x}^{\alpha }} = \mathop{\lim }\limits_{{t \rightarrow + \infty }}\frac{{t}^{\alpha ... | Yes |
\[ \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{{\log }_{a}x}{{x}^{\alpha }} = 0 \] for \( \alpha > 0 \), that is, for any positive exponent \( \alpha \) we have \( {\log }_{a}x = o\left( {x}^{\alpha }\right) \) as \( x \rightarrow + \infty \) . | Proof If \( a > 1 \), we set \( x = {a}^{t/\alpha } \) . Then by the properties of power functions and the logarithm, the theorem on the limit of a composite function, and the result of Example 29, we find \[ \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{{\log }_{a}x}{{x}^{\alpha }} = \mathop{\lim }\limits_{{t... | Yes |
Let us show further that \[ {x}^{\alpha }{\log }_{a}x = o\left( 1\right) \;\text{ as }x \rightarrow 0, x \in {\mathbb{R}}_{ + } \] for any \( \alpha > 0 \) . | Proof We need to show that \( \mathop{\lim }\limits_{{{\mathbb{R}}_{ + } \ni x \rightarrow 0}}{x}^{\alpha }{\log }_{a}x = 0 \) for \( \alpha > 0 \) . Setting \( x = 1/t \) and applying the theorem on the limit of a composite function and the result of the preceding example, we find \[ \mathop{\lim }\limits_{{{\mathbb{R... | Yes |
Example 34 The functions \( \left( {2 + \sin x}\right) x \) and \( x \) are of the same order as \( x \rightarrow \infty \), but \( \left( {1 + \sin x}\right) x \) and \( x \) are not of the same order as \( x \rightarrow \infty \) . | The condition that \( f \) and \( g \) be of the same order over the base \( \mathcal{B} \) is obviously equivalent to the condition that there exist \( {c}_{1} > 0 \) and \( {c}_{2} > 0 \) and an element \( B \in \mathcal{B} \) such that the relations\n\n\[ \n{c}_{1}\left| {g\left( x\right) }\right| \leq \left| {f\lef... | Yes |
Example 35 \( {x}^{2} + x = \left( {1 + \frac{1}{x}}\right) {x}^{2} \sim {x}^{2} \) as \( x \rightarrow \infty \) . | The absolute value of the difference of these functions\n\n\[ \left| {\left( {{x}^{2} + x}\right) - {x}^{2}}\right| = \left| x\right| \]\n\ntends to infinity. However, the relative error \( \frac{\left| x\right| }{{x}^{2}} = \frac{1}{\left| x\right| } \) that results from replacing \( {x}^{2} + x \) by the equivalent f... | Yes |
Let us show that \( \ln \left( {1 + x}\right) \sim x \) as \( x \rightarrow 0 \) . | \[ \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\ln \left( {1 + x}\right) }{x} = \mathop{\lim }\limits_{{x \rightarrow 0}}\ln {\left( 1 + x\right) }^{1/x} = \ln \left( {\mathop{\lim }\limits_{{x \rightarrow 0}}{\left( 1 + x\right) }^{1/x}}\right) = \ln \mathrm{e} = 1. \] Here we have used the relation \( {\log }_{a}\... | Yes |
Let us show that \( {\mathrm{e}}^{x} = 1 + x + o\left( x\right) \) as \( x \rightarrow 0 \) . | \[ \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{{\mathrm{e}}^{x} - 1}{x} = \mathop{\lim }\limits_{{t \rightarrow 0}}\frac{t}{\ln \left( {1 + t}\right) } = 1. \] Here we have made the substitution \( x = \ln \left( {1 + t}\right) ,{\mathrm{e}}^{x} - 1 = t \) and used the relations \( {\mathrm{e}}^{x} \rightarrow {\mat... | Yes |
Let us show that \( {\left( 1 + x\right) }^{\alpha } = 1 + {\alpha x} + o\left( x\right) \) as \( x \rightarrow 0 \) . | \[ \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{{\left( 1 + x\right) }^{\alpha } - 1}{x} = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{{\mathrm{e}}^{\alpha \ln \left( {1 + x}\right) } - 1}{\alpha \ln \left( {1 + x}\right) } \cdot \frac{\alpha \ln \left( {1 + x}\right) }{x} = \] \[ = \alpha \mathop{\lim }\limits_{{... | Yes |
Proposition 3 If \( f{ \sim }_{\mathcal{B}}\widetilde{f} \), then \( \mathop{\lim }\limits_{\mathcal{B}}f\left( x\right) g\left( x\right) = \mathop{\lim }\limits_{\mathcal{B}}\widetilde{f}\left( x\right) g\left( x\right) \), provided one of these limits exists. | Proof Indeed, given that \( f\left( x\right) = \gamma \left( x\right) \widetilde{f}\left( x\right) \) and \( \mathop{\lim }\limits_{\mathcal{B}}\gamma \left( x\right) = 1 \), we have\n\n\[ \mathop{\lim }\limits_{\mathcal{B}}f\left( x\right) g\left( x\right) = \mathop{\lim }\limits_{\mathcal{B}}\gamma \left( x\right) \w... | Yes |
\[ \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\ln \cos x}{\sin \left( {x}^{2}\right) } = \frac{1}{2}\mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\ln {\cos }^{2}x}{{x}^{2}} = \frac{1}{2}\mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\ln \left( {1 - {\sin }^{2}x}\right) }{{x}^{2}} = \] | \[ = \frac{1}{2}\mathop{\lim }\limits_{{x \rightarrow 0}}\frac{-{\sin }^{2}x}{{x}^{2}} = - \frac{1}{2}\mathop{\lim }\limits_{{x \rightarrow 0}}\frac{{x}^{2}}{{x}^{2}} = - \frac{1}{2}\text{.} \] Here we have used the relations \( \ln \left( {1 + \alpha }\right) \sim \alpha \) as \( \alpha \rightarrow 0,\sin x \sim x \) ... | Yes |
Example 42 \( \sqrt{{x}^{2} + x} \sim x \) as \( x \rightarrow + \infty \), but\n\n\[ \mathop{\lim }\limits_{{x \rightarrow + \infty }}\left( {\sqrt{{x}^{2} + x} - x}\right) \neq \mathop{\lim }\limits_{{x \rightarrow + \infty }}\left( {x - x}\right) = 0. \] | In fact,\n\n\[ \mathop{\lim }\limits_{{x \rightarrow + \infty }}\left( {\sqrt{{x}^{2} + x} - x}\right) = \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{x}{\sqrt{{x}^{2} + x} + x} = \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{1}{\sqrt{1 + \frac{1}{x}} + 1} = \frac{1}{2}. \] | Yes |
Proposition 4 For a given base\na) \( o\left( f\right) + o\left( f\right) = o\left( f\right) \) ; | Proof a) After the clarification just given, this assertion ceases to appear strange. The first symbol \( o\left( f\right) \) in it denotes a function of the form \( {\alpha }_{1}\left( x\right) f\left( x\right) \), where \( \mathop{\lim }\limits_{\mathcal{B}}{\alpha }_{1}\left( x\right) = 0 \) . The second symbol \( o... | Yes |
\[ \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{x - \sin x}{{x}^{3}} = ? \] | \[ \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{x - \sin x}{{x}^{3}} = \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{x - \left( {x - \frac{1}{3!}{x}^{3} + O\left( {x}^{5}\right) }\right) }{{x}^{3}} = \mathop{\lim }\limits_{{x \rightarrow 0}}\left( {\frac{1}{3!} + O\left( {x}^{2}\right) }\right) = \frac{1}{3!}. \] | Yes |
Example 44 Let us find\n\n\\[ \n\\mathop{\\lim }\\limits_{{x \\rightarrow \\infty }}{x}^{2}\\left( {\\sqrt[7]{\\frac{{x}^{3} + x}{1 + {x}^{3}}} - \\cos \\frac{1}{x}}\\right) .\n\\] | As \( x \\rightarrow \\infty \) we have:\n\n\\[ \n\\frac{{x}^{3} + x}{1 + {x}^{3}} = \\frac{1 + {x}^{-2}}{1 + {x}^{-3}} = \\left( {1 + \\frac{1}{{x}^{2}}}\\right) {\\left( 1 + \\frac{1}{{x}^{3}}}\\right) }^{-1} = \n\\]\n\n\\[ \n= \\left( {1 + \\frac{1}{{x}^{2}}}\\right) \\left( {1 - \\frac{1}{{x}^{3}} + O\\left( \\frac... | Yes |
\[ \mathop{\lim }\limits_{{x \rightarrow \infty }}{\left\lbrack \frac{1}{\mathrm{e}}{\left( 1 + \frac{1}{x}\right) }^{x}\right\rbrack }^{x} = \mathop{\lim }\limits_{{x \rightarrow \infty }}\exp \left\{ {x\left( {\ln {\left( 1 + \frac{1}{x}\right) }^{x} - 1}\right) }\right\} = \] | \[ = \mathop{\lim }\limits_{{x \rightarrow \infty }}\exp \left\{ {{x}^{2}\ln \left( {1 + \frac{1}{x}}\right) - x}\right\} = \] \[ = \mathop{\lim }\limits_{{x \rightarrow \infty }}\exp \left\{ {{x}^{2}\left( {\frac{1}{x} - \frac{1}{2{x}^{2}} + O\left( \frac{1}{{x}^{3}}\right) }\right) - x}\right\} = \] \[ = \mathop{\lim... | Yes |
If \( f : E \rightarrow \mathbb{R} \) is a constant function, then \( f \in C\left( E\right) \) . | This is obvious, since \( f\left( E\right) = c \subset V\left( c\right) \), for any neighborhood \( V\left( c\right) \) of \( c \in \mathbb{R} \) . | Yes |
The function \( f\left( x\right) = x \) is continuous on \( \mathbb{R} \) | Indeed, for any point \( {x}_{0} \in \mathbb{R} \) we have \( \left| {f\left( x\right) - f\left( {x}_{0}\right) }\right| = \left| {x - {x}_{0}}\right| < \varepsilon \) provided \( \left| {x - {x}_{0}}\right| < \delta = \varepsilon \) | Yes |
Example 3 The function \( f\left( x\right) = \sin x \) is continuous on \( \mathbb{R} \) . | In fact, for any point \( {x}_{0} \in \mathbb{R} \) we have\n\n\[ \left| {\sin x - \sin {x}_{0}}\right| = \left| {2\cos \frac{x + {x}_{0}}{2}\sin \frac{x - {x}_{0}}{2}}\right| \leq \]\n\n\[ \leq 2\left| {\sin \frac{x - {x}_{0}}{2}}\right| \leq 2\left| \frac{x - {x}_{0}}{2}\right| = \left| {x - {x}_{0}}\right| < \vareps... | Yes |
Example 4 The function \( f\left( x\right) = \cos x \) is continuous on \( \mathbb{R} \) . | Indeed, as in the preceding example, for any point \( {x}_{0} \in \mathbb{R} \) we have\n\n\[ \left| {\cos x - \cos {x}_{0}}\right| = \left| {-2\sin \frac{x + {x}_{0}}{2}\sin \frac{x - {x}_{0}}{2}}\right| \leq \]\n\n\[ \leq 2\left| {\sin \frac{x - {x}_{0}}{2}}\right| \leq \left| {x - {x}_{0}}\right| < \varepsilon \]\n\... | Yes |
The function \( f\left( x\right) = {a}^{x} \) is continuous on \( \mathbb{R} \) . | Indeed by property 3) of the exponential function (see Paragraph d in Sect. 3.2.2, Example 10a), at any point \( {x}_{0} \in \mathbb{R} \) we have\n\n\[ \mathop{\lim }\limits_{{x \rightarrow {x}_{0}}}{a}^{x} = {a}^{{x}_{0}} \]\n\nwhich, as we now know, is equivalent to the continuity of the function \( {a}^{x} \) at th... | Yes |
The function \( f\left( x\right) = {\log }_{a}x \) is continuous at any point \( {x}_{0} \) in its domain of definition \( {\mathbb{R}}_{ + } = \{ x \in \mathbb{R} \mid x > 0\} \) . | In fact, by property 3) of the logarithm (see Paragraph d in Sect. 3.2.2, Example 10b), at each point \( {x}_{0} \in {\mathbb{R}}_{ + } \) we have\n\n\[ \mathop{\lim }\limits_{{{\mathbb{R}}_{ + } \ni x \rightarrow {x}_{0}}}{\log }_{a}x = {\log }_{a}{x}_{0} \]\n\nwhich is equivalent to the continuity of the function \( ... | Yes |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.