Find $$\lim_{x\to0} \left(\cos x\right)^{\cot^2 x}$$
Sadly, I'm stuck trying to solve this. I'm assuming I have to use L'Hopital's rule, but I don't see how I can. It isn't homework or anything, just revising limits.
Any guidance is appreciated
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$\begingroup$ Have you tried using the rule $u^v = \exp( v \log u)$? $\endgroup$ – user49640 Jun 13 '17 at 14:29
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2$\begingroup$ Is it $\lim_{x\to 0} (\cos x)^{\cot(x^2)}$ or $\lim_{x\to 0}\cos(x^{\cot(x^2)})$ or something else? $\endgroup$ – Thomas Andrews Jun 13 '17 at 14:29
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$\begingroup$ (cos(x)) ^ ((cosx)^2) @ThomasAndrews $\endgroup$ – Stefanie Jun 13 '17 at 14:33
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1$\begingroup$ We can use WolframAlpha and see that the answer is $\frac 1{\sqrt{e}}$ $\endgroup$ – lioness99a Jun 13 '17 at 14:50
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1$\begingroup$ Sonnhard Graubner has written a hint. It's the same idea that was in my comment above. Would you like to try and take it from there? You can edit your calculations into your question and then we can comment. $\endgroup$ – user49640 Jun 13 '17 at 14:56
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HINT: your limit is $$e^{\lim_{x\to 0}\frac{\ln(cos(x))}{\tan(x)^2}}$$ and use L'Hopital
answered Jun 13 '17 at 14:37
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$\begingroup$ Could you please elaborate on how you you arrived at the hint? $\endgroup$ – Stefanie Jun 13 '17 at 15:00
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$\begingroup$ @Stefanie It's based on what I wrote in my comment above. $\endgroup$ – user49640 Jun 13 '17 at 15:01
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$\begingroup$ @Stefanie This is a common trick for then you have a complicated function as an exponent. Another way to look at this is this: Suppose the limit is $L$, so $\lim_{x\to0} \left(\cos x\right)^{\cot^2 x}=L$. Now take $\ln$ of both sides, $\ln(\lim_{x\to0} \left(\cos x\right)^{\cot^2 x}) = \ln L$. You can bring the limit on the outside of the $\ln$, and solve the limit. You should get $\ln ( \frac {1}{\sqrt e}) = \ln L$. Make each side an exponent of $e$, and you get $L = \dfrac {1}{\sqrt e}$ $\endgroup$ – Ovi Jun 18 '17 at 5:39
