Question

Find the value of ${lim}_{x\to 0}{\tan (\frac{\pi }{4}+x)}^{1/x}$.

Answer 1

$\underset{x\to }{lim}\tan (\frac{\pi }{4}+x{)}^{\frac{1}{x}}$

$\underset{x\to 0}{lim}\left({\tan }^{\frac{1}{x}}\right(\frac{\pi }{4}+x\left)\right)$

Refine: $\underset{r\to 0}{lim}\left[\tan \right(x+\frac{\pi }{4}{)}^{\frac{1}{x}}]$

Apply exponents rule : $a^x=e^{ln\left(ax\right)}=e^{x.ln\left(a\right)}$

$\left(\tan \right(x+\frac{\pi }{4}){)}^{\frac{1}{x}}=e^{\frac{1}{x}ln\left(\tan \right(x+\frac{\pi }{4}\left)\right)}$

$=\underset{x\to 0}{lim}\left(e^{\frac{1}{x}ln\left(\tan \right(x+\frac{\pi }{4}\left)\right)}\right)$

Apply limit chain rule i,e.

if $\underset{u\to 0}{lim}f\left(4\right)=L\&\underset{x\to 0}{lim}g\left(x\right)=b,\&f\left(x\right)$ is

continuous at $x=b$, then $\underset{x\to a}{lim}f\left(f\right(g\left(x\right))=L$

$\therefore g\left(x\right)=\frac{1}{2}\left(\tan \right(x+\frac{\pi }{4}\left)\right),f\left(4\right)=e^4$

$\therefore \underset{x\to 0}{lim}\left(\frac{1}{2}ln\right(\tan (x+\frac{\pi }{4})\left)\right)=2$

$\underset{u\to 2}{lim}\left(e^4\right)=e^2$

$u\to 2+$

By chain rule,

$=e^2$