# Evaluating the following limit without L'Hopital's help

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 $$e=\underset{x\to 0}{\mathop{\lim }}\,{{\left( {{1+x}} \right)}^{{1/x}}}$$

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

\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 $$e^{1/2}$$.
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 $$\sqrt e$$.