Calculation of the limit $\lim_{x\to 0} (\cos x)^{1/x^2}$ without De l'Hospital/Landau's symbols/asymptotic comparison I have calculate this limit 
$$\lim_{x\to 0}\ (\cos x)^{1/x^2}$$ 
with these steps. I have considered that:
$$(\cos x)^{1/x^2}=(\cos x -1+1)^{1/x^2}=\left(\frac{1}{\frac{1}{\cos x -1}}+1\right)^{\frac{1}{x^2}}$$
I remember that $1/(\cos x -1)$ when $x\to 0$ the limit is $\infty$. Hence 
$$\left(\frac{1}{\frac{1}{\cos x -1}}+1\right)^{\frac{1}{x^2}}=\left[\left(\frac{1}{\frac{1}{\cos x -1}}+1\right)^{\frac{1}{x^2}}\right]^{\frac{\frac{1}{\cos x -1}}{\frac{1}{\cos x -1}}}=\left[\left(\frac{1}{\frac{1}{\cos x -1}}+1\right)^{\frac{1}{\cos x-1}}\right]^{\frac{\frac{1}{x^2}}{\frac{1}{\cos x -1}}} \tag{1}$$
But if I take 
$$p=\frac{1}{\cos x -1}\xrightarrow{x\to 0}p\to \infty$$
therefore I consider the 
$$\lim_{p\to \infty}\left(1+\frac 1p\right)^p=e$$
Consequently for the $(1)$,
$$\left(\frac{1}{\frac{1}{\cos x -1}}+1\right)^{\frac{1}{\cos x-1}}\xrightarrow{p\to \infty} e$$
and the exponent
$$\lim_{x\to 0}\frac{\frac{1}{x^2}}{\frac{1}{-(-\cos x +1)}}=-\frac 12\tag{2}$$
At the end $\displaystyle \lim_{x\to 0}\ (\cos x)^{1/x^2}=e^{-\frac 12}$.
I have followed this strategy in my classroom with my students. Is there a shorter solution to the exercise than the one I have given?
 A: So you are assuming the fact $(1+u)^{1/u}\rightarrow e$ as $u\rightarrow 0$. With such, we also have $\dfrac{1}{u}\cdot\log(1+u)\rightarrow 1$, then
\begin{align*}
\dfrac{1}{x^{2}}\cdot\log(\cos x)&=\dfrac{1}{\cos x-1}\cdot\log(1+(\cos x-1))\cdot\dfrac{\cos x-1}{x^{2}}\\
&=\dfrac{1}{\cos x-1}\cdot\log(1+(\cos x-1))\cdot-2\cdot\dfrac{\sin^{2}\left(\dfrac{x}{2}\right)}{\left(\dfrac{x}{2}\right)^{2}}\cdot\dfrac{1}{4}\\
&\rightarrow-\dfrac{1}{2},
\end{align*}
so the limit goes to $e^{-1/2}$.
A: For all $x \in \mathbb{R}$ holds
$$1 - \frac{x^2}2\le \cos x \le 1 - \frac{x^2}2 + \frac{x^4}{24}$$
so
$$\left(1 - \frac{x^2}2\right)^{1/x^2}\le (\cos x)^{1/x^2} \le \left(1 - \frac{x^2}2 + \frac{x^4}{24}\right)^{1/x^2}$$
Both bounds are easily seen to converge to $e^{-\frac12}$ when $x \to 0$.
A: $$
\cos x = 1 - \frac{x^2} 2 + \frac{x^4}{24} - \frac{x^6}{720} + \cdots
$$
So for $x$ near $0$ we have
\begin{align}
(\cos x)^{1/x^2} & \ge \left( 1 - \frac{x^2} 2 \right)^{1/x^2} \\[10pt]
& = \left( 1 + \frac{-1/2}{u} \right)^u \\[8pt]
& \to e^{-1/2} \quad \text{as } u \to+\infty. \\[12pt]
(\cos x)^{1/x^2} & \le \left( 1 - \frac{x^2}{2+\varepsilon} \right)^{1/x^2} \text{ Why is this true? See below.} \\[8pt]
& = \left( 1 + \frac{-1/(2+\varepsilon)}{u} \right)^u \\[8pt]
& \to e^{-1/(2+\varepsilon)} \quad \text{as } u\to+\infty.
\end{align}
If the limit is $\ge e^{-1/2}$ and is $\le e^{-1/(2+\varepsilon)}$ for EVERY sufficiently small $\varepsilon>0,$ then the limit is $\le\lim_{\varepsilon\,\downarrow\,0} e^{-1/(2+\varepsilon)} = e^{-1/2}.$
$\text{“Why is this true? See below.''}$
$$
\cos x \le \underbrace{ 1 - \frac{x^2} 2 + \frac{x^4}{24} \le 1 - \frac{x^2}{2+\varepsilon} }
$$
The inequality over the $\underbrace{\text{underbrace}}$ holds whenever $x^2\le 12\varepsilon/(2+\varepsilon),$ and thus holds in the limit as $x\to0.$
A: With high school students I follow roughly the same approach. The only difference is that I originally propose them the fundamental limit as
$$(1+\alpha(x))^{\frac1{\alpha(x)}} {\to}\ \mbox{e},$$
when $\alpha(x) \to 0$. 
In this way I can avoid all the reciprocals you have in your expression. In your case, of course, $\alpha(x) = \cos x -1$, so that
\begin{eqnarray}
\lim_{x\to 0} \left\{\underbrace{[1+(\cos x-1)]^{\frac1{\cos x-1}}}_{\to \mbox{e}}\right\}^{\frac{\cos x -1}{x^2}}=\mbox{e}^{-\frac12}
\end{eqnarray}
