Mathematics - Limits Question with Solution | TestHub

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