l'hospital of a not defined function Good evening,
I would like to prove that: $$\lim_{x\to \frac{\pi}{2}} \frac{\tan(5x)}{\tan(x)} =\frac{1}{5}$$
But I can't figure out what kind of function it is in order to use l'hospital $$\frac{\infty}{\infty},\frac{0}{0}, ...$$
as $$\tan(\frac{\pi}{2})$$ is not defined.
I would be glad if you could give me a hint.
Best regards

Note that
$$\tan \left(\frac{\pi}{2}-\theta \right)=\frac{1}{\tan \theta}$$
thus let $x=\frac{\pi}{2}-y$ with $y\to0$
$$\frac{\tan 5x}{\tan x}=\frac{\tan\left(5\frac{\pi}{2}+5y\right)}{\tan(\frac{\pi}{2}+y)}=\frac{\tan\left(\frac{\pi}{2}+5y\right)}{\tan(\frac{\pi}{2}+y)}=\frac{\tan\left(y\right)}{\tan(5y)}=\frac{\tan y}{y}\frac{5y}{\tan(5y)}\cdot\frac15\to\frac15$$
Note that from the step
$$...=\frac{\tan y}{\tan(5y)}$$
you can also apply l'Hopital rule and obtain
$$\frac{\cos^2(5y}{\cos^2(y)}\cdot \frac15\to \frac11 \cdot \frac15 =\frac15$$
