Let f(x)=lim _ n → ∞ Σ_ =0^ n ( tan (x / 2^r+1 )+tan ^3 (x / 2^r+1 ) 1-tan ^2 (x / 2^r+1 ) ). Then lim _x → 0 e^x- e^f(x) (x-f(x)) is equal to

$\displaystyle \begin{aligned} & f(x)=\lim _{x \rightarrow \infty}\left(\frac{\tan \left(x / 2^{r+1}\right)+\tan ^3\left(x / 2^{r+1}\right)}{1-\tan ^2\left(x / 2^{r+1}\right)}\right) \\ & =\lim _{x \rightarrow \infty} \frac{\tan \left(\frac{x}{2^{r+1}}\right)}{\cos \left(\frac{x}{2^r}\right)} \\ & =\lim _{x \rightarrow \infty} \frac{\sin \left(x / 2^{r+1}\right)}{\cos \left(\frac{x}{2^{r+1}}\right) \cos \left(\frac{x}{2^r}\right)} \\ & =\lim _{x \rightarrow \infty} \frac{\sin \left(\frac{x}{2^r}-\frac{x}{2^{r+1}}\right)}{\cos \left(\frac{x}{2^{r+1}}\right) \cos \left(\frac{x}{2^r}\right)}\end{aligned}$​f(x)=x→∞lim​(1−tan2(x/2r+1)tan(x/2r+1)+tan3(x/2r+1)​)=x→∞lim​cos(2rx​)tan(2r+1x​)​=x→∞lim​cos(2r+1x​)cos(2rx​)sin(x/2r+1)​=x→∞lim​cos(2r+1x​)cos(2rx​)sin(2rx​−2r+1x​)​​ $\displaystyle =\lim _{x \rightarrow \infty} \tan \left(\frac{x}{2^r}\right)-\tan \left(\frac{x}{2^{r+1}}\right)$=x→∞lim​tan(2rx​)−tan(2r+1x​) From condition given question $\begin{aligned}
& \therefore \lim {n \rightarrow \infty} \sum{r=0}^n\left[\tan \left(\frac{x}{2^r}\right)-\tan \left(\frac{x}{2^{r+1}}\right)\right]=\tan x \
& \therefore \lim _{x \rightarrow 0}\left(\frac{e^x-e^{\tan x}}{x-\tan x}\right) \
& \Rightarrow \lim _{x \rightarrow 0} e^{\tan x}\left(\frac{e^{x-\tan x}-1}{x-\tan x}\right) \
& \Rightarrow \lim _{x \rightarrow 0} e^{\tan x} \lim _{x \rightarrow 0}\left(\frac{e^{x-\tan x}-1}{x-\tan x}\right) \
& \Rightarrow 1.1\left(\because \lim _{x \rightarrow 0} \frac{e^{x-1}}{x}=1\right) \
& \quad=1
\end{aligned}$