JEE Main 2025 — 28 January, Evening Shift — Question 22

Given the function:

$$f(x) = \lim_{n \to \infty} \sum_{r=0}^{n} \left( \frac{\tan(x/2^{r+1}) + \tan^3(x/2^{r+1})}{1 - \tan^2(x/2^{r+1})} \right)$$

We need to find the value of:

$$\lim_{x \to 0} \frac{e^x - e^{f(x)}}{x - f(x)}$$

Simplify the General Term Let $\theta = \frac{x}{2^{r+1}}$. The general term $T_r$ of the summation is:

$$T_r = \frac{\tan \theta + \tan^3 \theta}{1 - \tan^2 \theta} = \tan \theta \left( \frac{1 + \tan^2 \theta}{1 - \tan^2 \theta} \right)$$

Using the identity $\cos(2\theta) = \frac{1 - \tan^2 \theta}{1 + \tan^2 \theta}$, we have:

$$T_r = \tan \theta \cdot \sec(2\theta) = \frac{\tan \theta}{\cos(2\theta)}$$

Using the identity $\tan(2\theta) - \tan \theta = \frac{\tan \theta}{\cos(2\theta)}$, we can rewrite the term as:

$$T_r = \tan(2\theta) - \tan \theta = \tan\left(\frac{x}{2^r}\right) - \tan\left(\frac{x}{2^{r+1}}\right)$$

Evaluate the Summation The sum is a telescoping series:

$$S_n = \sum_{r=0}^{n} \left[ \tan\left(\frac{x}{2^r}\right) - \tan\left(\frac{x}{2^{r+1}}\right) \right]$$

Expanding the sum:

$$S_n = \left( \tan x - \tan \frac{x}{2} \right) + \left( \tan \frac{x}{2} - \tan \frac{x}{4} \right) + \dots + \left( \tan \frac{x}{2^n} - \tan \frac{x}{2^{n+1}} \right)$$

All intermediate terms cancel out, leaving:

$$S_n = \tan x - \tan\left(\frac{x}{2^{n+1}}\right)$$

As $n \to \infty$, $\frac{x}{2^{n+1}} \to 0$, so $\tan\left(\frac{x}{2^{n+1}}\right) \to 0$. Thus: $f(x) = \tan x$ Evaluate the Limit We now calculate the limit:

$$L = \lim_{x \to 0} \frac{e^x - e^{\tan x}}{x - \tan x}$$

Factor out $e^{\tan x}$ from the numerator:

$$L = \lim_{x \to 0} e^{\tan x} \left( \frac{e^{x - \tan x} - 1}{x - \tan x} \right)$$

Let $h = x - \tan x$. As $x \to 0$, $h \to 0$.

Using the standard limit $\lim_{h \to 0} \frac{e^h - 1}{h} = 1$:

$$L = e^{\tan(0)} \cdot 1 = e^0 \cdot 1 = 1$$

Final Answer: The value of the limit is $1$.