Limit evaluate $lim_{xto0}{{frac{ln(cos(4x))}{ln(cos(3x))}}}$?

now I am evaluating limits of functions, but i dont know how to start to solve this limit. It is possible without L Hopital's rule?

$\lim_{x\to0}{{\frac{\ln(\cos(4x))}{\ln(\cos(3x))}}}$?

Answer

One option (if you can use power series, which require at least as much calculus as L'Hopital's rule!):

In any sufficiently small neighborhood of $x = 0$, $\cos (ax) = \sqrt{1 - \sin^2(ax)}$. Thus the original quotient equals

$$\frac{\ln(1 - \sin^2(4x))}{\ln(1 - \sin^2(3x))} = \frac{ -\sin^2(4x) + O(x^4)}{-\sin^2(3x) + O(x^4)} = \frac{ -\sin^2(4x)/x^2 + O(x^2)}{-\sin^2(3x)/x^2 + O(x^2)} \to \frac{4^2}{3^2}$$