Question

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

then $\underset{x\to 0}{lim}\frac{f\left(x\right)-\sin x}{x^3}=?$

• A. $0$
• B. $-1$
• C. $\frac{1}{2}$
• D. $2$

Answer 1

Answer: (C)

$\frac{1}{2}$

$\tan x/2=\frac{\sin x/2}{\cos x/2}=\frac{\sin x}{1+\cos x}$

$=\frac{\sin x}{1+\cos x}=\frac{\frac{\sin x}{\cos x}}{\frac{1}{\cos x}+1}=\frac{\tan x}{\sec x+1}$

$\Rightarrow \tan \frac{x}{2}\sec x=\tan x-\tan \frac{x}{2}$

Replacing x by $\frac{x}{2}$

$\tan \left(\frac{x}{4}\right)\sec \frac{x}{2}=\tan \frac{x}{2}-\tan \frac{x}{4}$

$\tan \left(\frac{x}{2^{n+1}}\right)\sec \left(\frac{x}{2^n}\right)=\tan \left(\frac{x}{2^n}\right)-\tan \left(\frac{x}{2^{n+1}}\right)$

Repeating the steps we get

${\sum }_{r=0}^n\tan \frac{x}{2^{r+1}}\times \left(\frac{1+{\tan }^2\frac{x}{2^{r+1}}}{1-{\tan }^2\frac{x}{2^{r+1}}}\right)={\sum }_{r=0}^n\tan \frac{x}{2^{r+1}}\sec \frac{2x}{2^{r+1}}$

$={\sum }_{r=0}^n\tan \frac{x}{2^{r+1}}\sec \frac{x}{2^r}$

$=(\tan \left(x\right)-\tan \left(\frac{x}{2}\right))+(\tan \left(\frac{x}{2}\right)-\tan \left(\frac{x}{4}\right))+....+(\tan \left(\frac{x}{2^n}\right)-\tan \left(\frac{x}{2^{n+1}}\right))$

$=\tan x-\tan \left(\frac{x}{2^{n+1}}\right)$

$f\left(x\right)=\underset{n\to \infty }{lim}(\tan x-\tan \left(\frac{x}{2^{n+1}}\right))$

$=\tan x-0$

$=\tan x$

$\underset{x\to 0}{lim}\frac{f\left(x\right)-\sin x}{x^3}=\underset{x\to 0}{lim}\frac{\tan x-\sin x}{x^3}$

$=\underset{x\to 0}{lim}\frac{\tan x(1-\cos x)}{x^3}=\underset{x\to 0}{lim}\frac{\tan x}{x}\times \left(\frac{(1-\cos x)}{x^2}\right)$

$=1\times \frac{1}{2}=\frac{1}{2}$