Help finding $\lim_{x\to \pi/2} (\tan x)^{1/(\pi/2-x)}$

Question

can someone solve this it's probably de L'Hospital :( I spent all day and can do nothing. $$\lim_{x\to \pi/2} (\tan x)^{1/(\pi/2-x)}$$

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My work I try $\lim _{x\to \pi/2 }\left({\tan x}\right)^\frac{1}{\pi/2-x} =\lim _{x\to \pi/2 }\left(\frac{\sin x}{\cos x}\right)^\frac{1}{\pi/2-x} = \lim _{x\to\pi/2}$

1. $\displaystyle\lim _{x\to \pi/2 }(\sin x)^\frac{1}{\pi/2-x} = 1$

2. $\displaystyle\lim _{x\to \pi/2 }(\cos x)^\frac{1}{\pi/2-x} = \lim _{x\to \pi/2 }e^\frac{\ln\cos x}{\pi/2-x}$

Answer 1

Since $\tan(x)$, the base of the exponential, should be positive only the left limit is allowed. We have that $$\begin{align}\lim_{x\to \to (\pi/2)^-}{\left(\tan x\right)}^\frac{1}{\pi/2-x}&=\lim_{x\to \to (\pi/2)^-}\exp\left(\frac{\ln(\tan(x))}{\pi/2-x}\right)=+\infty \end{align}$$ because the argument of the exponential is not an indeterminate form (and we can not apply L'Hopital's rule) $$\frac{\ln(\tan(x))}{\pi/2-x}\to\frac{+\infty}{0^+}=+\infty.$$

P.S. Now are you able to find $\lim_{x\to (\pi/2)^-}(\cos x)^\frac{1}{\pi/2-x}$?

Answer 2

Without de Hôpital's rule.

First case:

$$\lim _{x\to \frac{\pi }{2}+}\left(\left(\tan \left(x\right)\right)^{\frac{1}{\frac{\pi }{2}-x}}\right)=*$$ Apply exponent rule $$a^x=e^{\ln \left(a^x\right)}=e^{x\cdot \ln \left(a\right)} \Longleftrightarrow \left(\tan \left(x\right)\right)^{\frac{1}{\frac{\pi }{2}-x}}=e^{\frac{1}{\frac{\pi }{2}-x}\ln \left(\tan \left(x\right)\right)}$$

$$*=\lim _{x\to \frac{\pi }{2}+}\left(e^{\frac{1}{\frac{\pi }{2}-x}\ln \left(\tan \left(x\right)\right)}\right)=**$$

But $$\lim _{x\to \frac{\pi }{2}+}\left(\frac{1}{\frac{\pi }{2}-x}\ln \left(\tan \left(x\right)\right)\right)=\lim _{x\to \frac{\pi }{2}+}\left(\frac{1}{\frac{\pi }{2}-x}\right)\cdot \lim \:_{x\to \frac{\pi }{2}+}\left(\ln \left(\tan \left(x\right)\right)\right)=-\infty\cdot +\infty=-\infty$$

being for $\:x\rightarrow\frac{\pi^+ }{2},\:x>\frac{\pi }{2}\quad \rightarrow\quad \frac{\pi }{2}-x<0$

Hence $$**=0$$

Second case:

$$\lim _{x\to \frac{\pi }{2}-}\left(\frac{1}{\frac{\pi }{2}-x}\ln \left(\tan \left(x\right)\right)\right)=\lim _{x\to \frac{\pi }{2}-}\left(\frac{1}{\frac{\pi }{2}-x}\right)\cdot \lim \:_{x\to \frac{\pi }{2}-}\left(\ln \left(\tan \left(x\right)\right)\right)=\infty\cdot\infty=+\infty$$ being for $\:x\rightarrow\frac{\pi^- }{2},\:x<\frac{\pi }{2}\quad \rightarrow \quad \frac{\pi }{2}-x>0$

Hence:

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