JEE Main 2025 — Limits Question with Solution

$\begin{array}{rl}& f(x)=\underset{x\to \infty }{\lim }\left(\frac{\tan \left(x/2^{r+1}\right)+{\tan }^3\left(x/2^{r+1}\right)}{1-{\tan }^2\left(x/2^{r+1}\right)}\right)\\ & =\underset{x\to \infty }{\lim }\frac{\tan \left(\frac{x}{2^{r+1}}\right)}{\cos \left(\frac{x}{2^r}\right)}\\ & =\underset{x\to \infty }{\lim }\frac{\sin \left(x/2^{r+1}\right)}{\cos \left(\frac{x}{2^{r+1}}\right)\cos \left(\frac{x}{2^r}\right)}\\ & =\underset{x\to \infty }{\lim }\frac{\sin \left(\frac{x}{2^r}-\frac{x}{2^{r+1}}\right)}{\cos \left(\frac{x}{2^{r+1}}\right)\cos \left(\frac{x}{2^r}\right)}\end{array}$ $={\lim }_{x\to \infty }\tan \left(\frac{x}{2^r}\right)-\tan \left(\frac{x}{2^{r+1}}\right)$ From condition given question $\begin{aligned} & \therefore \lim _{n \rightarrow \infty} \sum_{r=0}^n\left[\tan \left(\frac{x}{2^r}\right)-\tan \left(\frac{x}{2^{r+1}}\right)\right]=\tan x \\ & \therefore \lim _{x \rightarrow 0}\left(\frac{e^x-e^{\tan x}}{x-\tan x}\right) \\ & \Rightarrow \lim _{x \rightarrow 0} e^{\tan x}\left(\frac{e^{x-\tan x}-1}{x-\tan x}\right) \\ & \Rightarrow \lim _{x \rightarrow 0} e^{\tan x} \lim _{x \rightarrow 0}\left(\frac{e^{x-\tan x}-1}{x-\tan x}\right) \\ & \Rightarrow 1.1\left(\because \lim _{x \rightarrow 0} \frac{e^{x-1}}{x}=1\right) \\ & \quad=1 \end{aligned}$