Solve a limit without L'Hopital: $ \lim_{x\to0} \frac{\ln(\cos5x)}{\ln(\cos7x)}$ I need to solve this limit without using L'Hopital's rule. I have attempted to solve this countless times, and yet, I seem to always end up with an equation that's far more complicated than the one I've started with.
$$ \lim_{x\to0} \frac{\ln(\cos5x)}{\ln(\cos7x)}$$
Could anyone explain me how to solve this without using L'Hopital's rule ? Thanks!
 A: Hint use the facts that 
$$\frac{\ln x}{x-1}\to 1$$ and 
$$\frac{1-\cos x }{x^2}\to \frac{1}{2}$$
Merry Christmas.
A: HINT:
Note that we have  
$$\cos(n x)=1-2\sin^2(nx/2)$$
and 
$$ \log(\cos(nx))=\frac{\log(1-2\sin^2(nx/2)}{2\sin^2(nx/2)}\, \frac{2\sin^2(nx/2)}{(nx/2)^2} (nx/2)^2$$
A: $$x \to 0  \to \ln(1+x)\sim x\\ x \to 0 \to cos(ax)\sim1-\frac{1}{2}(ax)^2\\ 
\lim_{x \to 0}\frac{\ln (\cos 5x)}{\ln (\cos 7x)}=\\
\lim_{x \to 0}\frac{\ln (1-\frac{1}{2}(5x)^2)}{\ln (1-\frac{1}{2}(7x)^2)}=\\
\lim_{x \to 0}\frac{-\frac{1}{2}(5x)^2}{-\frac{1}{2}(7x)^2}=\\\frac{25}{49}$$
A: We have
$$\ln(\cos(X))=\ln(1+(\cos(X)-1))\sim \cos(X)-1\sim \frac{-X^2}{2}\;(X\to 0)$$
thus the limit is $$\frac{25}{49}.$$
A: You can use Cauchy's mean value theorem, but in fact that's kind of cheating, as this is how L' Hopital's rule is proved.
For every $x$ around 0 there exists by Cauchy's mean value theorem a $\vert \xi_x \vert \leq \vert x \vert$ such that
$$\frac{\ln(\cos(5x))}{\ln(\cos(7x))} 
= \frac{\ln(\cos(5x)) - \ln(\cos(0))}{\ln(\cos(7x))- \ln(\cos(0))}
= \frac{\left( \frac{-5\sin(5\xi_x)}{\cos(5\xi_x)} \right)}{\left( \frac{-7\sin(7\xi_x)}{\cos(7\xi_x)} \right)}
= \frac{5}{7}\frac{\cos(7\xi_x)}{\cos(5\xi_x)}\cdot \frac{\sin(5\xi_x)}{\sin(7\xi_x)}.$$
As $\xi_x\rightarrow 0$ for $x\rightarrow 0$, we get (if the respective limits exist)
$$ \lim_{x\rightarrow 0} \frac{\ln(\cos(5x))}{\ln(\cos(7x))} 
= \lim_{\xi \rightarrow 0} \frac{5}{7}\frac{\cos(7\xi_x)}{\cos(5\xi_x)}\cdot \frac{\sin(5\xi_x)}{\sin(7\xi_x)}
=\lim_{\xi \rightarrow 0} \frac{5}{7} \frac{\sin(5\xi_x)}{\sin(7\xi_x)}.$$
Repeating the argument for $\frac{\sin(5\xi_x)}{\sin(7\xi_x)}$ yields that the limit equals $\frac{25}{49}$.
