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I am given the following limit

$$\underset{x\to 0}{\mathop{\lim }}\,{{\left( {{cosx}} \right)}^{{-x}^{-2}}}$$

I tried using the fact that $cosx=1-2sin^2(\frac{x}{2})$, but it didn't give me anything. Moreover now I'm more confused because the answer is $\sqrt e$. Is this a typo or something? because I don't see how this limit can be equal to $\sqrt e$. Could you please help me evaluate this limit? Thank you for your help.

P.S. We didn't learn L'Hopital's rule, so I can't use it.

The only limit that I know that evaluates $e$ is $e=\underset{x\to 0}{\mathop{\lim }}\,{{\left( {{1+x}} \right)}^{{1/x}}}$

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  • $\begingroup$ Do you know Taylor polynomials? $\endgroup$
    – mathcounterexamples.net
    Apr 20, 2021 at 16:18
  • $\begingroup$ @mathcounterexamples.net no, we have just started evaluating limits of functions $\endgroup$
    – Britanica
    Apr 20, 2021 at 16:25

2 Answers 2

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\begin{align*} (\cos x)^{-x^{-2}}&=(1-(1-\cos x))^{\frac{1}{-(1-\cos x)}\cdot\frac{1-\cos x}{x^{2}}}, \end{align*} now \begin{align*} \dfrac{1-\cos x}{x^{2}}=\dfrac{1}{2}\left(\dfrac{\sin(\frac{x}{2})}{\frac{x}{2}}\right)^{2}\rightarrow\dfrac{1}{2}, \end{align*} so the limit tends to $e^{1/2}$.

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  • $\begingroup$ oh, now I understand thank you. I could have never thought of this solution. Thank you. $\endgroup$
    – Britanica
    Apr 20, 2021 at 16:40
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Taking the log of the expression, you're looking at the limit of $-\frac{\ln(\cos x))}{x^2}$ when $x\to0$.

Since $\cos x-1\sim -\frac{x^2}{2}$ and $\ln(1+x)\sim x$, you get by composition

$$-\frac{\ln(\cos x))}{x^2}\sim\frac12$$

Taking the exp finally yields $\sqrt e$.

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