I've seen some similar question, but none with $x\to\infty$ $$\lim_{x\to\infty}\frac{\ln\cos\frac {48} x}{\ln\cos\frac{1}{12x}}$$ one of the ideas that i tried is, with $t=\frac{48}{x}$, changing the limit to: $$\lim_{t\to0}\frac{\ln\cos t}{\ln\cos\frac{t}{576}}$$ but it got me stuck pretty much there.
Thanks and sorry if it's basically a duplicate.
In this answer, we reach the solution using the well-known inequality $\color{#c00}{e^x≥x+1}$ .
We prove the following limit, which is equivalent to the original limit :
$$\lim_{x\to 0^{+}}\frac {\ln \cos ax}{\ln \cos bx}=\frac {a^2}{b^2}$$
Using the substitution $x\longmapsto ±\ln x$ and applying the above inequality for $0<x<1$, we have :
$$ \begin{align}\ln x&≤x-1\\ \ln x&≥\frac {x-1}{x}\end{align} $$
and
$$ \begin{align}\frac {1}{\ln x}&≥\frac {1}{x-1}\\ \frac {1}{\ln x}&≤\frac {x}{x-1} \end{align} $$
The above inequalities lead to the following inequality :
$$ \begin{align}\cos bx\cdot\frac {1-\cos ax}{1-\cos bx}≤\frac{\ln\cos ax}{\ln\cos bx}≤\frac{1}{\cos ax}\cdot\frac {1-\cos ax}{1-\cos bx}\end{align} $$
where $0<\cos ax<1$ and $0<\cos bx<1$ .
Since
$$ \begin{align}\lim_{x\to 0}\frac{1-\cos ax}{1-\cos bx}=\frac{a^2}{b^2}\cdot\lim_{x\to 0}\frac{\frac{1-\cos ax}{(ax)^2}}{\frac{1-\cos bx}{(bx)^2}}=\frac{a^2}{b^2}\end{align} $$
Then, by Squeeze theorem you are done .
