JEE Main 2025 Jan 28 Shift 2, Mathematics Q22: Definite Integral as Limit of a Sum

Using the identity,

$\frac{\tan \theta +{\tan }^3\theta }{1-{\tan }^2\theta }=\tan 2\theta -\tan \theta$

where $\theta =\frac{x}{2^{r+1}}$,

$\therefore \frac{\tan \left(\frac{x}{2^{r+1}}\right)+{\tan }^3\left(\frac{x}{2^{r+1}}\right)}{1-{\tan }^2\left(\frac{x}{2^{r+1}}\right)}=\tan \left(\frac{x}{2^r}\right)-\tan \left(\frac{x}{2^{r+1}}\right)$

Hence,

$f(x)=\underset{n\to \infty }{\lim }\sum _{r=0}^n\left(\tan \frac{x}{2^r}-\tan \frac{x}{2^{r+1}}\right)=\tan x$

Now,

$\underset{x\to 0}{\lim }\frac{e^x-e^{f(x)}}{x-f(x)}=\underset{x\to 0}{\lim }\frac{e^x-e^{\tan x}}{x-\tan x}$

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

$\boxed{1}$