Calculation of the limit $\lim_{x\to 0} (\cos x)^{1/x^2}$ without De l'Hospital/Landau's symbols/asymptotic comparison

Question

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?

Answer 1

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 $e^{-1/2}$.

Answer 2

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 $x \to 0$.

Answer 3

$$ \cos x = 1 - \frac{x^2} 2 + \frac{x^4}{24} - \frac{x^6}{720} + \cdots $$ 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 $x\to0.$

Answer 4

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}