Limit evaluate $\lim_{x\to0}{{\frac{\ln(\cos(-5x))}{\ln(\cos(-3x))}}}$

Question

Please help me with this limit without using L'Hôpital's rule. I would by happy if you use simple solving. Thank you as much as I can ;).

$$\lim_{x\to0}{{\frac{\ln(\cos(-5x))}{\ln(\cos(-3x))}}}$$

Answer 1

We have the following Taylor series approximations about zero ($v=0$ and $w=0$):

$$\ln(1+w)=w+O(w^2)\sim w$$

and

$$\cos(v)=1-\tfrac{1}{2}v^2+O(v^4)\sim 1-\tfrac{1}{2}v^2$$

Numerator: Put $v=-5x$ to get $\cos(-5x)\sim1-\tfrac{25}{2}x^2$ and then put $w=-\tfrac{25}{2}x^2$ to get $\ln(\cos(-5x))\sim-\tfrac{25}{2}x^2$.

Denominator: Put $v=-3x$ to get $\cos(-3x)\sim1-\tfrac{9}{2}x^2$ and then put $w=-\tfrac{9}{2}x^2$ to get $\ln(\cos(-3x))\sim-\tfrac{9}{2}x^2$.

Hence, the limit is

$$L=\lim_{x\to0}\frac{\tfrac{25}{2}x^2}{\tfrac{9}{2}x^2}=\frac{25}{9}$$