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| # 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 | |