0
$\begingroup$

Can someone help me solve this limits? I know how to solve basic trigonometric limits, but this is too advanced for me. I know that i should start by writing that $\tan x=\sin x/\cos x$. Some instructions after that?

$$\Large\lim_{x\to\frac{\pi}{4}}(\tan x)^{\tan2x}$$

$$\Large\lim_{x\to\frac{\pi}{2}}(\tan x)^{\sin x}$$

$\endgroup$
  • $\begingroup$ If there are "strange" things in powers, you can always consider taking the logarithm, solving the limit and then taking the exponential of the result you get. $\endgroup$ – This Is Me Nov 15 '15 at 21:25
  • $\begingroup$ Yeah, but what do I do then? $\endgroup$ – Stefan Nov 15 '15 at 21:28
  • $\begingroup$ Then you basicly have the limit of a product, which is the product of the limits if both exist and otherwise you can usually apply l'Hopital in these situations. Note that both factors are basic goniometric limits (in the sense of not containing powers). $\endgroup$ – This Is Me Nov 15 '15 at 21:45
1
$\begingroup$

A limit of the form $L=\lim_{x\to a}f(x)^{g(x)}$ is most of the times treated by doing $$ \lim_{x\to a}g(x)\log f(x)=M $$ and so $L=e^M$ (including the cases of $M=-\infty$, where $L=0$, and $M=\infty$, where $L=\infty$).

In the first case you have to compute $$ \lim_{x\to\pi/4}\tan2x\log\tan x= \lim_{x\to\pi/4}\frac{\log\tan x}{\cot2x} $$ that's of the form $0/0$, so you can apply l'Hôpital: $$ \lim_{x\to\pi/4}\frac{\log\tan x}{\cot2x}= \lim_{x\to\pi/4}\frac{\dfrac{1}{\tan x}\dfrac{1}{\cos^2x}}{-\dfrac{2}{\sin^22x}}= \lim_{x\to\pi/4}\frac{\cos x}{\sin x}\frac{1}{\cos^2x}\frac{-4\sin^2x\cos^2x}{2} $$ that should have no problem, now.

It's similar for the second one.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.