Can't solve a difficult limit

Hint:

First the expression in two, and use equivalents:

We have $\cosh x\sim_{+\infty}\frac12\mathrm e^x$, so $1+\cosh x\sim \frac12\mathrm e^x$, and finally $$\ln(1+\cosh x)\sim_{+\infty}x-\ln 2\sim_{+\infty} x$$

. On the other hand, $$x(\sqrt{x^2 + x} - x)=\frac{x(\not x^2 + x - \not x^2)}{\sqrt{x^2 + x} + x}\sim_{+\infty}\frac{x^2}{2x}=\frac x2,$$ so that $$\frac{x(\sqrt{x^2 + x} - x)}{\ln(1+\cosh x)}\sim_{+\infty}\frac{\frac12 x}{x}=\frac 12.$$ Can you show that $$\frac{\cos x\ln x}{\ln(1+\cosh x)}\to 0?$$

Hint: Note that, for large $x$, $\cosh x$ behaves as $\frac12e^x$. So, $\lim_{x\to\infty}\frac{\log(\cosh x)}x=1$. And your limit is equal to$$\lim_{x\to\infty}\frac{\left(\sqrt{x^2+x}-x\right)+\frac{\cos(x)\log(x)}x}{\frac{\log(\cosh x)}x}.$$