If $f\left(x\right)=\underset{n\to \mathrm{\infty }}{lim} \sum _{r=0}^n \frac{\tan \frac{x}{2^r+1}+{\tan }^3\frac{x}{2^{r+1}}}{1-{\tan }^2\frac{x}{2^{r+1}}}$ then $\underset{x\to 0}{lim} \frac{f\left(x\right)}{x}$ is equal to

Let ${\alpha }_r=\frac{x}{2^{r+1}},r=0,1,2,\dots ,n.$ Then,

    $\begin{array}{l} \frac{\tan \frac{x}{2^{r+1}}+{\tan }^3\frac{x}{2^{r+1}}}{1-{\tan }^2\frac{x}{2^{r+1}}}\\ =\frac{\tan \alpha +{\tan }^3{\alpha }_r}{1-{\tan }^2{\alpha }_r}\\ =\tan {\alpha }_r\left(\frac{1+{\tan }^2{\alpha }_r}{1-{\tan }^2{\alpha }_r}\right)\end{array}$

     $\begin{array}{l}=\frac{\tan {\alpha }_r}{\cos 2{\alpha }_r}=\frac{\sin {\alpha }_r}{\cos {\alpha }_r\cos 2{\alpha }_r}\\ =\frac{\sin \left(2{\alpha }_r-{\alpha }_r\right)}{\cos {\alpha }_r\cos 2{\alpha }_r}=\tan 2{\alpha }_r-\tan {\alpha }_r\end{array}$

$\therefore f\left(x\right)=\underset{n\to \mathrm{\infty }}{lim} \sum _{r=0}^n \left(\tan 2{\alpha }_r-\tan {\alpha }_r\right)$

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

Hence, $\underset{x\to 0}{lim} \frac{f\left(x\right)}{x}=\underset{x\to 0}{lim} \frac{\tan x}{x}=1$