Find $\lim_{x \to 0}\left(\frac{\cos x}{\cos(2x)}\right)^{\frac{1}{x^2}}$ $$\lim_{x \to 0}\left(\frac{\cos x}{\cos(2x)}\right)^{\frac{1}{x^2}}$$
What I have done is to take the $\ln$ 
$$e^{\lim_{x \to 0}\ln\left(\left(\frac{\cos x}{\cos(2x)}\right)^{\frac{1}{x^2}}\right)}$$
$$y={\lim_{x \to 0}{\frac{1}{x^2}}\cdot \ln\frac{\cos x}{\cos(2x)}}$$
Applying l'hopital to get 
$$y={\lim_{x \to 0}{\frac{-2}{x^3}}\cdot \frac{\cos(2x)}{\cos x}\cdot \frac{-\sin x\cos(2x)+2\sin(2x)\cos x}{(\cos(2x)^2)}}={\lim_{x \to 0}\frac{-2}{x^3\cdot \cos x}\cdot \frac{-\sin x\cos(2x)+2\sin(2x)\cos x}{(\cos(2x))}}$$
Should I apply l'hopital again?

We can proceed using the expansion $\cos(x)=1-\frac12 x^2+O(x^4)$.  
First note that
$$\frac{\cos(x)}{\cos(2x)}=1+\frac32 x^2+O(x^4)$$
so that the limit of interest becomes 
$$\begin{align}
\lim_{x\to \infty}\left(\frac{\cos(x)}{\cos(2x)}\right)^{1/x^2}&=\lim_{x\to \infty}\left(\left(1+\frac{3/2}{1/x^2}\right)^{1/x^2}\left(1+O(x^4)\right)^{1/x^2}\right) \tag 1\\\\
\end{align}$$
Second, recalling that the limit definition of the exponential function is given by
$$e^x=\lim_{n\to \infty}\left(1+\frac xn\right)^n$$
we find that 
$$\lim_{x\to \infty} \left(1+\frac{3/2}{1/x^2}\right)^{1/x^2}=e^{\frac32} \tag 2$$
Third, using the expansion for the logarithm function $\log(1+x)=O(x)$, we find that 
$$\begin{align}
\lim_{x\to \infty}\left(1+O(x^4)\right)^{1/x^2}&=\lim_{x\to \infty}e^{\frac{1}{x^2}\log\left(1+O(x^4)\right)}\\\\
&=\lim_{x\to \infty}e^{O(x^2)}\\\\
&=1 \tag 3
\end{align}$$ 
Putting $(1)-(3)$ together yields the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\left(\frac{\cos(x)}{\cos(2x)}\right)^{1/x^2}=e^{3/2}}$$
