JEE (Main) Mathematics Limits and Derivatives

Limit Of Function Formulas

Limit of a function \( f(x) \) is said to exist as \( x \to a \) when,
\( \text{Limit}_{h \to 0^+} f(a - h) = \text{Limit}_{h \to 0^+} f(a + h) = \text{some finite value M.} \)
(Left hand limit) = (Right hand limit)

Indeterminate Forms:
\( \frac{0}{0}, \frac{\infty}{\infty}, 0 \times \infty, \infty - \infty, \infty^0, 0^0, \text{ and } 1^\infty \).

Standard Limits:
\( \text{Limit}_{x \to 0} \frac{\sin x}{x} = \text{Limit}_{x \to 0} \frac{\tan x}{x} = \text{Limit}_{x \to 0} \frac{\tan^{-1} x}{x} = \text{Limit}_{x \to 0} \frac{\sin^{-1} x}{x} = \text{Limit}_{x \to 0} \frac{e^x - 1}{x} = \text{Limit}_{x \to 0} \frac{\ln(1+x)}{x} = 1 \)
\( \text{Limit}_{x \to 0} (1 + x)^{1/x} = \text{Limit}_{x \to \infty} \left(1 + \frac{1}{x}\right)^x = e \), \( \text{Limit}_{x \to 0} \frac{a^x - 1}{x} = \log_e a, a > 0 \), \( \text{Limit}_{x \to a} \frac{x^n - a^n}{x - a} = na^{n-1} \).

Limits Using Expansion:
(i) \( a^x = 1 + \frac{x \ln a}{1!} + \frac{x^2 \ln^2 a}{2!} + \frac{x^3 \ln^3 a}{3!} + ........., a > 0 \)
(ii) \( e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + ...... \)
(iii) \( \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ........ \text{for } -1 < x \le 1 \)
(iv) \( \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ..... \)
(v) \( \cos x = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + ..... \)
(vi) \( \tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} + ...... \)
(vii) \( \tan^{-1} x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + .... \)
(viii) \( \sin^{-1} x = x + \frac{1^2}{3!} x^3 + \frac{1^2 \cdot 3^2}{5!} x^5 + \frac{1^2 \cdot 3^2 \cdot 5^2}{7!} x^7 + ..... \)
(x) for \( |x| < 1, n \in R (1 + x)^n = 1 + nx + \frac{n(n-1)}{1 \cdot 2} x^2 + \frac{n(n-1)(n-2)}{1 \cdot 2 \cdot 3} x^3 + ........... \)

Limits of form \( 1^\infty, 0^0, \infty^0 \):
Also for \( (1)^\infty \) type of problems we can use following rules.
\( \text{lim}_{x \to 0} (1 + x)^{1/x} = e, \text{lim}_{x \to a} [f(x)]^{g(x)}, \text{where } f(x) \to 1 ; g(x) \to \infty \text{ as } x \to a = \text{lim}_{x \to a} e^{\text{lim}_{x \to a} [f(x)-1] g(x)} \)

Sandwich Theorem or Squeeze Play Theorem:
If \( f(x) \le g(x) \le h(x) \forall x \) & \( \text{Limit}_{x \to a} f(x) = \ell = \text{Limit}_{x \to a} h(x) \) then \( \text{Limit}_{x \to a} g(x) = \ell \).