Vrftjkry

I have just started learning L'Hopital rule, and so far I thought I understood everything until I stumbled upon this question
$$lim_{xto 0} frac{ln(cos(ax))}{ln(cos(bx))}.$$

To this, eventually got
$$lim_{xto 0} frac{a sin(ax) cos(bx)}{b sin(bx)cos(ax)}$$

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?

You can apply l'Hôpital once more:
$$
lim_{xto 0}frac{a}{b}
frac{acos(ax)cos(bx)-bsin(ax)sin(bx)}
{bcos(bx)cos(ax)-asin(bx)sin(ax)}
$$
and now direct substitution is possible. Not the best method, however: you can observe that
$$
lim_{xto0}frac{cos(bx)}{cos(ax)}=1
$$
and reduce to computing
$$
lim_{xto0}frac{sin(ax)}{sin(bx)}
$$
which is $a/b$.

With Taylor expansion, note that, assuming $ane0$,
$$
cos(ax)=1-frac{(ax)^2}{2!}+o(x^2)qquad ln(1+t)=t+o(t)
$$
so that
$$
ln(cos(ax))=-frac{a^2x^2}{2}+o(x^2)
$$
Hence your limit is
$$
lim_{xto0}frac{-a^2x^2/2+o(x^2)}{-b^2x^2/2+o(x^2)}=frac{a^2}{b^2}
$$

$$lim_{xto}dfrac{sin ax}{sin bx}cdotdfrac{cos bx}{cos ax}$$

$$=dfrac ablim_{xto0}dfrac{sin ax}{ax}cdotlim_{xto0}dfrac{bx}{sin bx}lim_{xto0}dfrac{cos bx}{cos ax}=?$$

Mind you $lim_{xto0}implies xne0$

Using $cos^2y=1-sin^2y,$

$$2lim_{xto0}dfrac{ln(cos ax)}{x^2}=-a^2lim_{xto0}dfrac{ln(1-sin^2ax)}{-sin^2ax}left(lim_{xto0}dfrac{sin(ax)}{ax}right)^2=-a^2$$

From here by standard limits

$$ldots=lim_{xto 0} frac{a sin(ax) cos(bx)}{b sin(bx)cos(ax)}=lim_{xto 0} frac a bfrac{tan (ax)}{tan (bx)}=lim_{xto 0} frac {a^2} {b^2}frac{tan (ax)}{ax}frac{bx}{tan (bx)}=frac {a^2} {b^2}cdot 1 cdot 1 =frac {a^2} {b^2}$$