Do not understand L'Hopital Rule 
I have just started learning L'Hopital rule, and so far I thought I understood everything until I stumbled upon this question 
  $$\lim_{x\to 0}  \frac{\ln(\cos(ax))}{\ln(\cos(bx))}.$$

To this, eventually got
$$\lim_{x\to 0}  \frac{a \sin(ax) \cos(bx)}{b \sin(bx)\cos(ax)}$$
From my knowledge, $\sin(0)$ is 0!! and the whole thing will be ''$\frac{0}{0}$'', however the answer key I was given does not continue the implementation of the L'Hopital rule, but instead obtains the answer $\frac{a^2}{b^2}$.
Is there some important concept I'm missing out? Or is the differentiation supposed to continue and the answer key just skipped the steps?
 A: $$\lim_{x\to}\dfrac{\sin ax}{\sin bx}\cdot\dfrac{\cos bx}{\cos ax}$$
$$=\dfrac ab\lim_{x\to0}\dfrac{\sin ax}{ax}\cdot\lim_{x\to0}\dfrac{bx}{\sin bx}\lim_{x\to0}\dfrac{\cos bx}{\cos ax}=?$$
Mind you $\lim_{x\to0}\implies x\ne0$
A: You can apply l'Hôpital once more:
$$
\lim_{x\to 0}\frac{a}{b}
\frac{a\cos(ax)\cos(bx)-b\sin(ax)\sin(bx)}
     {b\cos(bx)\cos(ax)-a\sin(bx)\sin(ax)}
$$
and now direct substitution is possible. Not the best method, however: you can observe that
$$
\lim_{x\to0}\frac{\cos(bx)}{\cos(ax)}=1
$$
and reduce to computing
$$
\lim_{x\to0}\frac{\sin(ax)}{\sin(bx)}
$$
which is $a/b$.
With Taylor expansion, note that, assuming $a\ne0$,
$$
\cos(ax)=1-\frac{(ax)^2}{2!}+o(x^2)\qquad \ln(1+t)=t+o(t)
$$
so that
$$
\ln(\cos(ax))=-\frac{a^2x^2}{2}+o(x^2)
$$
Hence your limit is
$$
\lim_{x\to0}\frac{-a^2x^2/2+o(x^2)}{-b^2x^2/2+o(x^2)}=\frac{a^2}{b^2}
$$
A: Using $\cos^2y=1-\sin^2y,$
$$2\lim_{x\to0}\dfrac{\ln(\cos ax)}{x^2}=-a^2\lim_{x\to0}\dfrac{\ln(1-\sin^2ax)}{-\sin^2ax}\left(\lim_{x\to0}\dfrac{\sin(ax)}{ax}\right)^2=-a^2$$
A: From here by standard limits
$$\ldots=\lim_{x\to 0}  \frac{a \sin(ax) \cos(bx)}{b \sin(bx)\cos(ax)}=\lim_{x\to 0} \frac a b\frac{\tan (ax)}{\tan (bx)}=\lim_{x\to 0} \frac {a^2} {b^2}\frac{\tan (ax)}{ax}\frac{bx}{\tan (bx)}=\frac {a^2} {b^2}\cdot 1 \cdot 1 =\frac {a^2} {b^2}$$
