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?
 A: Setting $x=2y$
$$\lim_{y\to0}\left(1+\dfrac{\cos2y-\cos4y}{\cos4y}\right)^{1/4y^2}$$
$$=\left[\lim_{y\to0}\left(1+\dfrac{2\sin3y\sin y}{\cos4y}\right)^{\frac{\cos4y}{2\sin3y\sin y}}\right]^{\lim_{y\to0}\frac{2\sin3y\sin y}{4y^2\cos4y}}$$
The inner limit converges to $e$
The exponent 
$$\lim_{y\to0}\frac{2\sin3y\sin y}{4y^2\cos4y}=\dfrac32\cdot\dfrac1{\lim_{y\to0}\cos4y}\cdot\lim_{y\to0}\dfrac{\sin3y}{3y}\cdot\lim_{y\to0}\dfrac{\sin y}y=?$$
A: If $L=\lim\ (\frac{\cos\ x}{\cos\ 2x})^\frac{1}{x^2}$ then
\begin{align} \ln\ L &= \lim \frac{\ln\ \cos\ x -\ln\ \cos\ 2x}{x^2}
\\
&=\lim \ \frac{ \frac{-\sin\ x}{\cos\ x} + \frac{2\sin\ 2x}{\cos\
2x}}{2x} \\&= \lim\ \frac{-\sin\ x +2\sin\ 2x}{2x}\\& = \lim\
\frac{-\cos\ x + 4\cos\ 2x}{2} =\frac{3}{2},\ L=e^\frac{3}{2}
\end{align}
A: Shortly, developing a minimum of terms,
$$\left(\frac{\cos(x)}{\cos(2x)}\right)^{1/x^2}
\approx\left(\frac{1-\dfrac{x^2}2}{1-2x^2}\right)^{1/x^2}
\approx\left(1+\frac32x^2\right)^{1/x^2}$$ which tends to $e^{3/2}$.
A: If you start from $g(x) = \frac{\cos x}{\cos(2x)}^{\frac{1}{x^2}}$ and first take logs you have $\ln g(x) = {\frac{1}{x^2}} (\ln \frac{\cos x}{\cos(2x)})$
Expanding $\frac{\cos(x)}{\cos(2x)} \approx 1 + 3/2 x^2$ and $\ln (1+x) \approx  x$ you have $\ln g(x)  \approx \frac{3/2 x^2}{x^2} = 3/2$, so that the limit is $e^{3/2}$.
A: 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
$$\bbox[5px,border:2px solid #C0A000]{\lim_{x\to 0}\left(\frac{\cos(x)}{\cos(2x)}\right)^{1/x^2}=e^{3/2}}$$
