Limits problem without L'Hopital

Question

I am prompted to solve the following limit

$$\lim_{x\mapsto 0} (\cos(x)^\frac{1}{x^2})$$

I try to approach this problem by doing

$$\lim_{x\mapsto 0} (-1+(1+\cos(x))^\frac{\cos(x)}{\cos(x)x^2})$$

where the limit of ($1+\cos(x)^\frac{1}{\cos(x)}$) is $e$. So I would have to calcualte the limit of $\frac{\cos(x)}{x^2}$. Is this approach correct?

Answer 1

I would say:

$\displaystyle \lim_{x\to 0} \cos(x)^\frac{1}{x^2}=\lim_{x\to 0}e^{\frac{1}{x^2}\ln \cos x}$

$\displaystyle \lim_{x\to 0} \frac{1}{x^2}\ln \cos x = \lim_{x\to 0} \frac{1}{x^2}\ln \sqrt{1-\sin^2x}=\frac{1}{2}\lim_{x\to 0} \frac{\sin ^{2}x}{x^{2}}\frac{\ln (1-\sin ^{2}x)}{\sin ^{2}x}=$

$\displaystyle \frac{1}{2}\lim_{x\to 0}\left(\frac{\sin x}{x}\right)^2\lim_{y\to 0}\frac{\ln (1-y)}{y}=\frac{1}{2}\cdot 1 \cdot (-1) = -\frac{1}{2}$

$\displaystyle \Rightarrow \lim_{x\to 0} \cos(x)^\frac{1}{x^2}=e^{-\frac{1}{2}}=\frac{1}{\sqrt{e}}$

Answer 2

$$(\cos x)^{1/x^2}=(1-\sin^2x)^{1/2x^2}$$

$$=\left[\lim_{\sin x\to0}\left(1-\sin^2x\right)^{(-1/\sin^2x)}\right]^{-\dfrac12(\lim_{x\to0}\sin x/x)^2}$$

Now use $$\lim_{h\to0}\left(1+h\right)^{1/h}=\lim_{n\to\infty}\left(1+\dfrac1n\right)^n=e$$ and $$\lim_{u\to0}\dfrac{\sin u}u=1$$