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

Question

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 1

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

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

Answer 2

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 $\large\color{red}{\frac{16}{9}}$.

Answer 3

Without Taylor or L'Hopital: We know $\lim_{u\to 0} [\ln (1+u)]/u = 1.$ (This is just $\ln'(1)=1.$) Thus

$$\ln (\cos u) = \frac{\ln (1+(\cos u-1))}{\cos u -1}(\cos u -1)$$ $$ =\frac{\ln (1+(\cos u-1))}{\cos u -1}\frac{1}{\cos u +1}(\cos^2 u -1)$$ $$ = \frac{\ln (1+(\cos u-1))}{\cos u -1}\frac{1}{\cos u +1}(-\sin^2 u)$$

As $u \to 0,$ this looks like $1\cdot (1/2)\cdot (-\sin^2 u).$ Use this on both $\ln (\cos 4x), \ln (\cos 3x).$

Answer 4

With equivalents:

$\ln(u)\sim_1 u-1$, hence $\enspace\dfrac{\ln\cos4x}{\ln\cos3x}\sim_0\dfrac{\cos4x-1}{\cos3x-1}\sim_0\dfrac{-\dfrac{16x^2}2}{-\dfrac{9x^2}2}=\dfrac{16}9. $