# Limits

## Standard Limits
- $\lim_{x \to 0} \frac{\sin x}{x} = 1$
- $\lim_{x \to 0} \frac{\tan x}{x} = 1$
- $\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}$
- $\lim_{x \to 0} \frac{e^x - 1}{x} = 1$
- $\lim_{x \to 0} \frac{\ln(1+x)}{x} = 1$
- $\lim_{x \to 0} \frac{a^x - 1}{x} = \ln a$
- $\lim_{x \to 0} (1 + x)^{1/x} = e$
- $\lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e$
- $\lim_{x \to 0} \frac{(1+x)^n - 1}{x} = n$

## L'Hôpital's Rule
If $\lim \frac{f(x)}{g(x)}$ gives $\frac{0}{0}$ or $\frac{\infty}{\infty}$:
$\lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}$ (if RHS exists)

Can be applied repeatedly if indeterminate form persists.

## Squeeze Theorem
If $g(x) \leq f(x) \leq h(x)$ near $a$, and $\lim_{x \to a} g(x) = \lim_{x \to a} h(x) = L$, then $\lim_{x \to a} f(x) = L$

## Indeterminate Forms
$\frac{0}{0}$, $\frac{\infty}{\infty}$, $0 \cdot \infty$, $\infty - \infty$, $0^0$, $\infty^0$, $1^\infty$

For $1^\infty$ form: $\lim [f(x)]^{g(x)} = e^{\lim g(x)[f(x)-1]}$ when $f(x) \to 1$, $g(x) \to \infty$

## JEE Tips
- Always check if direct substitution works first
- For $\frac{0}{0}$: try factoring, rationalizing, or L'Hôpital's
- For $1^\infty$: use the exponential limit formula
- Taylor series expansion useful for complex limits