Calculation of Trigonometric Limit with Summation. If $\displaystyle f(x)=\lim_{n\rightarrow \infty}\sum_{r=0}^{n}\left\{\frac{\tan \left(\frac{x}{2^{r+1}}\right)+\tan^3\left(\frac{x}{2^{r+1}}\right)}{1-\tan^2\left(\frac{x}{2^{r+1}}\right)}\right\}.$ Then  value of $f(x)$ is 
$\bf{My\; Try::}$ Let $\displaystyle \left(\frac{x}{2^{r+1}}\right)=y$. So Trigonometric Expression is
$\displaystyle \frac{\tan y\cdot \left(1+\tan^2 y\right)}{\left(1-\tan^2 y\right)}=\frac{\tan y}{\cos 2y}=\frac{\sin y}{\cos 2y \cdot \cos y} = \frac{1}{2}\left\{\frac{\sin (2y-y)}{\cos 2y \cdot \cos y}\right\} = \frac{1}{2}\left\{\tan (2y)-\tan (y)\right\}$
Now How Can I solve after that
Help me
Thanks
 A: You have the right start, but I prefer if you write $y_r = \frac{x}{2^{r+1}}$ so that $y_0 = \frac{x}{2}$ and you are trying to find 
$$
\frac{1}{2} \sum_{r=0}^{\infty} \left( \tan(2y_r) - \tan(y_r)\right)
$$
Since $y_r = 2y_{r+1}$, this becomes
$$
\frac{1}{2} \sum_{r=0}^{\infty} \left(\tan(2y_r) - \tan(2y_{r+1})\right)= \frac{1}{2}
\left[  \left( \tan(2y_0) - \tan(2y_1) \right) + \left( \tan(2y_1) - \tan(2y_2) \right) + \left( \tan(2y_2) - \tan(2y_3) \right) + \ldots \right]
$$
From this it is obvious that all terms except the first cancel, and since the limit for large $r$ of $\tan (2y_r)$ is zero, the sum is just the first term.
So the answer is
$$ 
\frac{1}{2} \tan (2y_0) = \frac{1}{2} \tan(x)
$$
A: $\displaystyle 2f(x)=2\sum_{r=0}^{n}\left\{\frac{\tan \left(\frac{x}{2^{r+1}}\right)+\tan^3\left(\frac{x}{2^{r+1}}\right)}{1-\tan^2\left(\frac{x}{2^{r+1}}\right)}\right\}.$
$\displaystyle=\sum_{r=0}^{n}\left\{\tan \frac x{2^r}-\tan \frac x{2^{r+1}}\right\}$ which is telescopic 
A: Continuing on from your answer,
$$\eqalign{\sum_{r=0}^{n}\left\{\frac{\tan \left(\frac{x}{2^{r+1}}\right)+\tan^3\left(\frac{x}{2^{r+1}}\right)}{1-\tan^2\left(\frac{x}{2^{r+1}}\right)}\right\}
  &=\frac{1}{2}\sum_{r=0}^{n}\Bigl(\tan\Bigl(\frac{x}{2^r}\Bigr)-\tan\Bigl(\frac{x}{2^{r+1}}\Bigr)\Bigr)\cr}$$
which is a telescoping sum, giving
$$\frac{1}{2}\left(\tan x-\tan\Bigl(\frac{x}{2}\Bigr)+\tan\Bigl(\frac{x}{2}\Bigr)-\tan\Bigl(\frac{x}{2^2}\Bigr)+\cdots-\tan\Bigl(\frac{x}{2^{n+1}}\Bigr)\right)\ .$$
The sum is
$$\frac{1}{2}\left(\tan x-\tan\Bigl(\frac{x}{2^{n+1}}\Bigr)\right)$$
which tends to $\frac{1}{2}\tan x$ as $n\to\infty$.
