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Corollary 12.1 The set function \( {\bar{c}}_{p}\left( \cdot \right) \) is countably sub-additive. | Proof By Proposition 11.1 the outer \( p \) -capacity is finitely sub-additive. Let \( \left\{ {E}_{n}\right\} \) be a countable collection of sets in \( {\mathbb{R}}^{N} \) . It suffices to consider the case when each \( {E}_{n} \) and \( \bigcup {E}_{n} \) are of finite outer \( p \) -capacity. The collection of sets... | Yes |
Proposition 15.1 \( {\mathcal{H}}^{N - s}\left( {E}_{s}\right) = 0 \) . | Proof For \( t > 0 \) introduce the set\n\n\[ \n{E}_{s, t} = \left\{ {x \in {\mathbb{R}}^{N}\left| {\mathop{\limsup }\limits_{{\rho \rightarrow 0}}{\rho }^{s}{f}_{{B}_{\rho }\left( x\right) }}\right| u \mid {dy} > t}\right\} .\n\]\n\nSince \( {E}_{s, t} \subset {E}_{s} \), one also has \( \mu \left( {E}_{s, t}\right) =... | Yes |
Lemma 2 Let \( h \) be an \( {\left( \mathcal{S} \times \mathfrak{I}\right) }^{ * } \) -measurable function on \( X \times Y \) such that \( h = 0 \) a.e. \( \left\lbrack {\mu \times \lambda }\right\rbrack \) . Then for almost all \( x \) & \( X \) it is true that \( h\left( {x, y}\right) = 0 \) for almost all \( y \) ... | PROOF OF LEMMA 2 Let \( P \) be the set of all points in \( X \times Y \) at which \( h\left( {x, y}\right) \neq 0 \) . Then \( {P\varepsilon }{\left( \mathcal{S} \times \mathfrak{I}\right) }^{ * } \) and \( \left( {\mu \times \lambda }\right) \left( P\right) = 0 \) . Hence there exists a \( {Q\varepsilon }\mathcal{S} ... | Yes |
Lemma 1 If \( \mu \) is a positive or real Borel measure on \( {R}^{k} \), then \( \bar{D}\mu \) is a Borel function. | PROOF If \( \alpha \) is a real number and \( {\bar{\Delta }}_{r}\left( x\right) > \alpha \) for some \( {x\varepsilon }{R}^{k} \) and some \( r > 0 \) (the notation is as in Definition 8.3), then there exists an \( E \) & \( \Omega \) such that \( {x\varepsilon E} \), diam \( E < r \), and \( \mu \left( E\right) > {\a... | Yes |
Lemma 2 Suppose \( \mu \) is a positive Borel measure on \( {R}^{k} \) which is finite on compact sets. Let \( A \) be a Borel set for which \( \mu \left( A\right) = 0 \) . Then \( \left( {D\mu }\right) \left( x\right) = 0 \) a.e. \( \left\lbrack m\right\rbrack \) on \( A \) . | PROOF If \( P \) is the set of all \( x \) at which \( \left( {\bar{D}\mu }\right) \left( x\right) > 0 \), Lemma 1 shows that \( P \) is a Borel set, and hence so is \( A \cap P \) . We have to prove that \( m\left( {A \cap P}\right) = 0 \) .\n\nAssume this is false. Then there exists an \( \alpha > 0 \) and a Borel se... | Yes |
Lemma 3 If \( \mu \bot m \), then \( \left( {D\mu }\right) \left( x\right) = 0 \) a.e. \( \left\lbrack m\right\rbrack \) . | PROOF It is enough to prove this for real \( \mu \) . In that case \( \mu = {\mu }^{ + } - {\mu }^{ - } \) (Jordan decomposition theorem), where \( {\mu }^{ + } \geq 0,{\mu }^{ + } \bot m \), and similar statements apply to \( {\mu }^{ - } \) . Since \( {\mu }^{ + } \bot m \), there is a Borel set \( A \) such that \( ... | Yes |
Theorem 9.2(d) shows that \( \widehat{g} = {\left| \widehat{f}\right| }^{2} \geq 0 \), and since \( H\left( {\lambda t}\right) \) increases to 1 as \( \lambda \rightarrow 0 \), the monotone convergence theorem gives | \[ \mathop{\lim }\limits_{{\lambda \rightarrow 0}}{\int }_{-\infty }^{\infty }H\left( {\lambda t}\right) \widehat{g}\left( t\right) {dm}\left( t\right) = {\int }_{-\infty }^{\infty }{\left| \widehat{f}\left( t\right) \right| }^{2}{dm}\left( t\right) . \] Now (5),(6), and (7) show that \( {f\varepsilon }{L}^{2} \) and t... | Yes |
Theorem 14.18(b) shows that \( f \) is one-to-one on \( \bar{\Omega } \) . | Since every continuous one-to-one mapping of a compact set has a continuous inverse ([26], Theorem 4.17), the proof is complete. | Yes |
Corollary 2 The mapping \( x \rightarrow {x}^{-1} \) is differentiable. Its differential at any \( x \) & \( G \) is the linear operator which takes \( h \) & \( A \) to \( - {x}^{-1}h{x}^{-1} \) . | This can also be read off from (1). Note that the notion of the differential of a transformation makes sense in any normed linear space, not just in \( {R}^{k} \), as in Definition 8.22. If \( A \) is commutative, the above differential takes \( h \) to \( - {x}^{-2}h \), which agrees with the fact that the derivative ... | No |
Corollary 3 For every \( {x\varepsilon A},\sigma \left( x\right) \) is compact, and \( \left| \lambda \right| \leq \parallel x\parallel \) if \( {\lambda \varepsilon \sigma }\left( x\right) \) . | For if \( \left| \lambda \right| > \parallel x\parallel \), then \( e - {\lambda }^{-1}{x\varepsilon G} \), by Theorem 18.3, and the same is true of \( x - {\lambda e} = - \lambda \left( {e - {\lambda }^{-1}x}\right) \) ; hence \( \lambda \neq \sigma \left( x\right) \) . To prove that \( \sigma \left( x\right) \) is cl... | Yes |
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 \( F \) : \... | 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 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 closed, that is, no external forces are acting o... | Yes |
Example 13. The diagonal\n\n\[ \n\Delta = \left\{ {\left( {a, b}\right) \in {X}^{2} \mid a = b}\right\} \n\]\n\nis 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 |
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, b}\rig... | It is known from geometry that the relation of parallelism between lines has the following properties:\n\n\( a\mathcal{R}a \) (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).\n... | Yes |
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 \( b \) ; ... | 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\[ a\mathcal{R}b \mathrel{\text{:=}} \left( {a \subset b}\right) .\n\]\n\nThis relation obviously has the following properties:\n\n\( a\mathcal{R}a \) (reflexivity);... | Yes |
Example 3. \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\left( {1 + \frac{{\left( -1\right) }^{n}}{n}}\right) = 1 \) | since \( \left| {\left( {1 + \frac{{\left( -1\right) }^{n}}{n}}\right) - 1}\right| = \frac{1}{n} < \varepsilon \) when \( n > \left\lbrack \frac{1}{\varepsilon }\right\rbrack \) . | Yes |
Example 6. 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. | Proof. 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 \( \varepsilon = \frac{\left| A\right| }{2} > 0 \)... | Yes |
An ultimately constant sequence converges. | If \( {x}_{n} = A \) for \( n > N \), then for any neighborhood \( V\left( A\right) \) of \( A \) we have \( {x}_{n} \in V\left( A\right) \) when \( n > N \), that is, \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{x}_{n} = A \) . | Yes |
Theorem 3. a) Let \( \\left\\{ {x}_{n}\\right\\} \) and \( \\left\\{ {y}_{n}\\right\\} \) be two convergent sequences with \( \\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}{x}_{n} = A \) and \( \\mathop{\\lim }\\limits_{{n \\rightarrow \\infty }}{y}_{n} = B \) . If \( A < B \), then there exists an index \( N \\i... | Proof. a) Choose a number \( C \) such that \( A < C < B \) . By definition of limit, we can find numbers \( {N}^{\\prime } \) and \( {N}^{\\prime \\prime } \) such that \( \\left| {{x}_{n} - A}\\right| < C - A \) for all \( n > {N}^{\\prime } \) and \( \\left| {{y}_{n} - B}\\right| < B - C \) for all \( n > {N}^{\\pri... | 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 \( \left| {{x}_{n} - A}\right| < \frac{\varepsilon }{2} \) for \( n > N \) . Then if \( m > N \) and \( n > N \), we have \( \left| {{x}_{m} - {x}_{n}}\right| \leq \) \( \lef... | Yes |
The sequence \( {\left( -1\right) }^{n}\left( {n = 1,2,\ldots }\right) \) has no limit, since it is not a Cauchy sequence. | The negation of the statement that \( \left\{ {x}_{n}\right\} \) is a Cauchy sequence is the following:\n\n\[ \exists \varepsilon > 0\;\forall N \in \mathbb{N}\;\exists n > N\;\exists m > N\;\left( {\left| {{x}_{m} - {x}_{n}}\right| \geq \varepsilon }\right) \;,\]\n\nthat is, there exists \( \varepsilon > 0 \) such tha... | Yes |
Example 9. Let\n\n\\[ \n{x}_{1} = 0,\;{x}_{2} = 0.{\alpha }_{1},\;{x}_{3} = 0.{\alpha }_{1}{\alpha }_{2},\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 by adjoining a 0 or a 1 to its predecessor. W... | 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 |
Consider the sequence \( \left\{ {x}_{n}\right\} \), where\n\n\[ \n{x}_{n} = 1 + \frac{1}{2} + \cdots + \frac{1}{n} \n\] | Since\n\[ \n\left| {{x}_{2n} - {x}_{n}}\right| = \frac{1}{n + 1} + \cdots + \frac{1}{n + n} > n \cdot \frac{1}{2n} = \frac{1}{2}, \n\]\n\nfor all \( n \in \mathbb{N} \), the Cauchy criterion implies immediately that this sequence does not have a limit. | Yes |
In order for a nondecreasing sequence to have a limit it is necessary and sufficient that it be bounded above. | 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 }\... | 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... | 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 \( n! \mathrel{\text{:=}} \) \( 1 \cdot 2 \cdot \ldots \cdot n \) . | 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 is not... | Yes |
Let us prove that the limit \( \mathop{\lim }\limits_{{n \rightarrow \infty }}{\left( 1 + \frac{1}{n}\right) }^{n} \) exists. | We begin by verifying the following inequality, sometimes called Jakob Bernoulli’s inequality: \n\n\[ \n{\left( 1 + \alpha \right) }^{n} \geq 1 + {n\alpha }\text{for}n \in \mathbb{N}\text{and}\alpha > - 1\text{.} \n\] \n\nProof. The assertion is true for \( n = 1 \) . If it holds for \( n \in \mathbb{N} \), then it mus... | 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}}\right\... | 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... | No |
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}{l} - \frac{1}{n},\text{ if }n = {2m} + 1 \\ - \f... | 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 \i... | 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 \( \mathop{\lim }\limits_{{k \rightarrow \infty }}{x}_{k} = \mathop{\lim }\limits_{{k \rightarrow \infty }}{x}_{k} = + \infty \) or \( \mathop{\lim }\limits_{{k \rightarrow \infty }}{x}_{k} = \mathop{\lim }\limits_{{k \rightarrow \infty }}{x}_{k} = \) \( - \infty \) have been investigated above, a... | 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 s... | 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 \( s = \mathop{\lim }\limits_{{k \rightarrow \infty }}{x}_{k} \) are finite and, by what has been proved, are partial limits of the sequence. Only when \( i = s \... | 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 + ... | Yes |
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 t... | 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\lef... | Yes |
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 \( \alpha = \) \( \mathop{\lim }\limits_{{n \rightarrow \infty }}\sqrt[n]{\left| {a}_{n}\right| } \) . Then the following are true:\n\na) if \( \alpha < 1 \), the series \( \mathop{\sum }\limits_{{n = 1}}^{\in... | 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 \( \alpha = \overline{\mathop{\lim }\limits_{{n \rightarrow \infty }}}\;\sqrt[n]{\left| \frac{{\left( 2 + {\left( -1\right) }^{n}\right) }^{n}}{{x}^{n}}\right| } = \left| x\right| \;\overline{\mathop{\lim }\limits_{{n \rightarrow \infty }}}\;\left| {2 + {\left( -1\right) }^{n}}\right| = 3\left| x\right| . \)... | Yes |
Corollary 10. (d’Alembert’s test). \( {}^{6} \) 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 convergen... | 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\n\[ \n{2}^{n}{a}_{{2}^{n + 1}} \leq {a}_{{2}^{n} + 1} + \cdots + {a}_{{2}^{n + 1}} \leq {2}... | Yes |
Let \( E = \mathbb{R} \smallsetminus 0 \), and \( f\left( x\right) = x\sin \frac{1}{x} \). We shall verify that\n\n\[ \mathop{\lim }\limits_{{E \ni x \rightarrow 0}}x\sin \frac{1}{x} = 0 \] | 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 \( \overset{ \circ }{U}\left( 0\right) \) of 0 . Hence for any neighborhood \( V\left( 1\right) \) we obtain \( f\left( {\overset{ \circ... | Yes |
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 \( {\left. \operatorname{sgn}\right| }_{{\mathbb{R}}_{ - }} \) of sgn to \( {\mathbb{R}}_{ - } \) is a constant function equal to -1 and \( {\left... | \[ \mathop{\lim }\limits_{{{\mathbb{R}}_{ - } \ni x \rightarrow 0}}\operatorname{sgn}x = - 1,\text{ and }\mathop{\lim }\limits_{{{\mathbb{R}}_{ + } \ni x \rightarrow 0}}\operatorname{sgn}x = 1, \] | 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 \( U\left( 0\right) \) 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 cannot both lie in the ... | 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\le... | 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\na) \( \math... | As already noted at the beginning of Subsect. 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... | No |
Proposition 2. a) If \( \alpha : E \rightarrow \mathbb{R} \) and \( \beta : E \rightarrow \mathbb{R} \) are infinitesimal functions as \( E \ni x \rightarrow a \), then their sum \( \alpha + \beta : E \rightarrow \mathbb{R} \) is also infinitesimal as \( E \ni x \rightarrow a \) . | a) We shall verify that\n\n\[ \left( {\mathop{\lim }\limits_{{E \ni x \rightarrow a}}\alpha \left( x\right) = 0}\right) \land \left( {\mathop{\lim }\limits_{{E \ni x \rightarrow a}}\beta \left( x\right) = 0}\right) \Rightarrow \left( {\mathop{\lim }\limits_{{E \ni x \rightarrow a}}\left( {\alpha + \beta }\right) \left(... | Yes |
Theorem 3. a) If the functions \( f : E \rightarrow \mathbb{R} \) and \( g : E \rightarrow \mathbb{R} \) are such that \( \mathop{\lim }\limits_{{E \ni x \rightarrow a}}f\left( x\right) = A \), and \( \mathop{\lim }\limits_{{E \ni x \rightarrow a}}g\left( x\right) = B \) and \( A < B \), then there exists a deleted nei... | Proof. a) Choose a number \( C \) such that \( A < C < B \) . By definition of limit, we find deleted neighborhoods \( {\overset{ \circ }{{U}^{\prime }}}_{E}\left( a\right) \) and \( {\overset{ \circ }{{U}^{\prime \prime }}}_{E}\left( a\right) \) of \( a \) in \( E \) such that \( \left| {f\left( x\right) - A}\right| <... | Yes |
\[ \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\sin x}{x} = 1 \] | 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 exists \( B \in \mathcal{B} \) such t... | 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... | 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 \... | 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) = \mathop{... | 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 mappi... | Yes |
We shall show that\n\n\[ \mathop{\lim }\limits_{{t \rightarrow 0}}{\left( 1 + t\right) }^{1/t} = \mathrm{e}. \]\n | 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 3 of Sect. 3.1, we can consider the auxiliary mapping \( f : {\mathbb{R}}_{ + } \rightarrow \mathbb{N} \) given by the function \( \left\lbrack x\right\rbrack ... | 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 \( \mathop{\lim }\limits_{{E \ni x \rightarrow s}}f\left( x\right) \) .\n\nIf this limit exists, then, like any function having a limit, the function \( f \) is ultimately bounded over the base \( E \ni x \rightarrow s \) .\n\nSince \( f \) is nondecreasing on \( E \), it follo... | Yes |
Example 24. \( {x}^{2} = o\left( x\right) \) as \( x \rightarrow 0 \) | since \( {x}^{2} = x \cdot x \) | No |
Example 25. \( x = o\left( {x}^{2}\right) \) as \( x \rightarrow \infty \) | since ultimately (as long as \( x \neq 0 \) ), \( x = \frac{1}{x} \cdot {x}^{2} \) | 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}^{\... | 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 |
Let us show that\n\n\[ \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{{\log }_{a}x}{{x}^{\alpha }} = 0 \] \n\nfor \( \alpha > 0 \), 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\n\n\[ \mathop{\lim }\limits_{{x \rightarrow + \infty }}\frac{{\log }_{a}x}{{x}^{\alpha }} = \mathop{\lim }\limits... | 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{... | Yes |
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\[\n\left| {\left( {{x}^{2} + x}\right) - {x}^{2}}\right| = \left| x\right|\n\]\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... | Yes |
Example 38. Let us show that \( \ln \left( {1 + x}\right) \sim x \) as \( x \rightarrow 0 \) . | Proof.\n\n\[ \mathop{\lim }\limits_{{x \rightarrow 0}}\frac{\ln \left( {1 + x}\right) }{x} = \mathop{\lim }\limits_{{x \rightarrow 0}}\ln {\left( 1 + x\right) }^{1/x} = \ln \left( {\mathop{\lim }\limits_{{x \rightarrow 0}}{\left( 1 + x\right) }^{1/x}}\right) = \ln \mathrm{e} = 1. \]\n\nHere we have used the relation \(... | 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 {\math... | 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 \( f\underset{\mathcal{B}}{ \sim }\widetilde{f} \), 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) \... | 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 \(... | 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 |
Let us find\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) . | As \( x \rightarrow \infty \) we have:\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\[ = \left( {1 + \frac{1}{{x}^{2}}}\right) \left( {1 - \frac{1}{{x}^{3}} + O\left( \frac{1}{{x}^{6}}\right) }\right... | Yes |
Example 45.\n\n\\[ \n\\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\... | \n\\[ \n= \\mathop{\\lim }\\limits_{{x \\rightarrow \\infty }}\\exp \\left\\{ {{x}^{2}\\ln \\left( {1 + \\frac{1}{x}}\\right) - x}\\right\\} =\n\\]\n\n\\[ \n= \\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) }\\... | 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 \( c \in \mathbb{R} \) . | No |
The function \( f\left( x\right) = x \) is continuous on \( \mathbb{R} \). | Indeed, for any point \( {x}_{0} \in \widetilde{\mathbb{R}} \) 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 |
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