Question

Solve for k:$\begin{array}{c}f\left(x\right)={\left[\tan \left(\frac{\pi }{4}+x\right)\right]}^{\frac{1}{x}},forx\ne 0\\ =k,forx=0\end{array},$ $\}atx=0$

Answer 1

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

$=k,x=0$

$\underset{x\to 0}{lim}f\left(x\right)=\underset{x\to 0}{lim}\left[tan\right(\frac{\pi }{4}+x){]}^{1/x}|1\infty$ from

$\underset{x\to 0}{lim}\left[tan\right(\frac{\pi }{4}+x)-1]\frac{1}{x}|\frac{0}{0}$ form

$=e$

$=e^{\underset{x\to 0}{lim}}\frac{se{c}^2(\frac{\pi }{4}+x)}{1}|$ L'Hospital's Rule

$=e^2$

f(x) is contentious of x = 0 only if $f\left(0\right)=e^2$

$\Rightarrow k=e^2$

$\therefore$ f(x) is contentious af x = 0, when $k=e^2$