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

By using Chebyshev polynomials of the first kind,

$$ \lim_{x\to 0}\frac{\log(\cos(4x))}{\log(\cos(3x))}=\lim_{t\to 1}\frac{\log(8t^3-8t^2+1)}{\log(4t^3-3t)}=\lim_{z\to 0}\frac{\log(1+16 z+40 z^2+32 z^3+8 z^4)}{\log(1 + 9 z + 12 z^2 + 4 z^3)}$$ hence the wanted limit equals $\large\color{red}{\frac{16}{9}}$.

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

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