Discussion Forum

If the function  $f(x) =\left[ \tan \left(\frac{\pi}{4}+x\right)\right]^{1/x}$    for x≠0 is K for x=0 continuous at x =0, then K = ?

Answer:

Explanation:

We have , $f(x) =\left[ \tan \left(\frac{\pi}{4}+x\right)\right]^{1/x}$ =K

 Since , f(x)  is continuous at x=0, then

 $f(o)=\lim_{x \rightarrow 0}f(x)$

  $=\lim_{x \rightarrow 0}\left[ \tan \left(\frac{\pi}{4}+x\right)\right]^{1/x}$

$\Rightarrow K=\lim_{x \rightarrow 0}\left[\frac{1+\tan x}{1-\tan x}\right]^{1/x}$    [$1^{\infty}$ form]

$=e^{\lim_{x \rightarrow 0}\left[\frac{1+\tan x}{1-\tan x}-1\right].\frac{1}{x}}$

   $=e^{\lim_{x \rightarrow 0}\left[\frac{2\tan x}{1-\tan x}\right].\frac{1}{x}}$

 $=e^{2\lim_{x \rightarrow 0}\frac{\tan x}{x}.\lim_{x \rightarrow 0}\frac{1}{1-\tan x}}$

                                     $\left[\because \lim_{x\rightarrow 0}\frac{\tan x}{x}=1\right]$

 $\therefore$      $K=e^{2.1(\frac{1}{1-0})}=e^{2}$