JEE Main · 2025 · 28 Jan 2025 · Shift 2 · Limits

Let $\theta =x/2^{r+1}$. Using $\tan 2\theta -\tan \theta =\frac{2\tan \theta }{1-{\tan }^2\theta }-\tan \theta =\frac{\tan \theta +{\tan }^3\theta }{1-{\tan }^2\theta }$, each term of the sum is


Summing from $r=0$ to $n$ telescopes to $\tan x-\tan \left(\frac{x}{2^{n+1}}\right)$, and as $n\to \infty$ this tends to $\tan x$. So $f(x)=\tan x$.

For the limit,


As $x\to 0$, $\tan x\to 0$ and $x-\tan x\to 0$, so the second factor tends to 1 (using $e^u-1\sim u$) and $e^{\tan x}\to 1$.

The limit equals 1.