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
$$\frac{\cos^2(5y}{\cos^2(y)}\cdot \frac15\to \frac11 \cdot \frac15 =\frac15$$
Hint: only use l'Hopital after first simplifying the expression:
$$\lim_{x\to \frac{\pi}{2}} \frac{\tan(5x)}{\tan(x)} = \lim_{x\to \frac{\pi}{2}} \frac{\;\;\cfrac{\sin(5x)}{\cos(5x)}\;\;}{\;\;\cfrac{\sin(x)}{\cos(x)}\;\;} = \left(\lim_{x\to \frac{\pi}{2}} \frac{\cos(x)}{\cos(5x)} \right) \cdot \left(\lim_{x\to \frac{\pi}{2}} \frac{\sin(5x)}{\sin(x)} \right)=\lim_{x\to \frac{\pi}{2}} \frac{\cos(x)}{\cos(5x)}$$
Put $u=\pi/2-x$, then $\tan(x)=\cot(u)=1/\tan(u)$. Also $\tan(5x)=\cot(5u)=1/\tan(5u)$ because $\tan(5/2-5u)=\tan(\pi/2-5u)$. Then with $u\rightarrow 0$ you can use L'Hospital's Rule.
