Question

If $f\left(x\right)=\{\begin{array}{c}{\left[\tan (\frac{\pi }{4}+x)\right]}^{\frac{1}{x}}\text{for}x\ne 0\\ \\ k\text{for}x=0\end{array}$ is continuous at $x=0$, find $k$.

Answer 1

The function would be continuous if ${lim}_{x\to 0}f\left(x\right)=f\left(0\right)$
${lim}_{x\to 0}f\left(x\right)=f\left(0\right)$
${lim}_{x\to 0}{\left[\tan (\frac{\pi }{4}+x)\right]}^{\frac{1}{x}}=k$
${lim}_{x\to 0}{\left[\frac{1+\tan x}{1-\tan x}\right]}^{\frac{1}{x}}=k$
${lim}_{x\to 0}{[1+\frac{1+\tan x}{1-\tan x}-1]}^{\frac{1}{x}}=k$
${lim}_{x\to 0}{[1+\frac{1+\tan x-1+\tan x}{1-\tan x}]}^{\frac{1}{x}}=k$
${lim}_{x\to 0}{[1+\frac{2\tan x}{1-\tan x}]}^{\frac{1}{x}}=k$
${lim}_{x\to 0}{[1+\frac{2\tan x}{1-\tan x}]}^{\frac{1}{\frac{2\tan x}{1-\tan x}}\times \frac{2\tan x}{x(1-\tan x)}}=k$

We know that,
$\underset{x\to 0}{lim}[1+x{]}^{\frac{1}{x}}=e$

$\therefore e^{{lim}_{x\to 0}2\frac{\tan x}{x(1-x)}}=k$
$e^{{lim}_{2x\to 0}\frac{\tan x}{x}}\underset{x\to 0}{\overset{\frac{1}{(1-\tan x)}}{lim}}=k[\because \underset{x\to 0}{lim}\left(\frac{\tan x}{x}\right)=1]$
$e^{2\times 1\times 1}=k$
$\therefore k=e^2$