Limits, Continuity and Differentiability – Continuity – JEE Main 24 Jan 2026 Shift 1

Solution:

For the function to be continuous at $x=0$, we must have:
$$f(0) = \lim_{x \to 0} f(x)$$

$$f(0) = \lim_{x \to 0} \frac{e^{\tan x} – e^{x} + \ln(\sec x + \tan x) – x}{\tan x – x}$$

This is a $\frac{0}{0}$ form. Using L’Hospital’s Rule, we differentiate the numerator and denominator with respect to $x$:
$$f(0) = \lim_{x \to 0} \frac{e^{\tan x} \sec^2 x – e^{x} + \sec x – 1}{\sec^2 x – 1}$$

We know that $\sec^2 x – 1 = \tan^2 x$. Rearranging the numerator to form standard limits:
$$f(0) = \lim_{x \to 0} \frac{e^{\tan x}(\sec^2 x – 1) + (e^{\tan x} – e^{x}) + (\sec x – 1)}{\tan^2 x}$$

Separating the terms:
$$f(0) = \lim_{x \to 0} \left( \frac{e^{\tan x}\tan^2 x}{\tan^2 x} + \frac{e^{x}(e^{\tan x – x} – 1)}{\tan^2 x} + \frac{\sec x – 1}{\tan^2 x} \right)$$

Evaluating each limit separately:
1. $\lim_{x \to 0} e^{\tan x} = e^0 = 1$

2. $\lim_{x \to 0} \frac{e^{x}(e^{\tan x – x} – 1)}{\tan^2 x}$
Using expansions, $\tan x – x \approx \frac{x^3}{3}$ and $\tan^2 x \approx x^2$. The order of the numerator is higher ($x^3$) than the denominator ($x^2$), so this limit is $0$.

3. $\lim_{x \to 0} \frac{\sec x – 1}{\tan^2 x} = \lim_{x \to 0} \frac{1 – \cos x}{\cos x \cdot \frac{\sin^2 x}{\cos^2 x}} = \lim_{x \to 0} \frac{1 – \cos x}{\sin^2 x} \cdot \cos x$
$$= \lim_{x \to 0} \frac{1 – \cos x}{(1 – \cos x)(1 + \cos x)} \cdot \cos x = \frac{1}{1+1} \cdot 1 = \frac{1}{2}$$

Adding them up:
$$f(0) = 1 + 0 + \frac{1}{2} = \frac{3}{2}$$

Ans. (4)