Another combined limit I've tried to get rid of those logarithms, but still, no result has came.
$$\lim_{x\to 0 x \gt 0} \frac{\ln(x+ \sqrt{x^2+1})}{\ln{(\cos{x})}}$$
Please help

It can be computed this way
$$
\frac{\ln (x+\sqrt{x^{2}+1})}{\ln {(\cos {x})}}=\frac{\ln (1+(x-1)+\sqrt{%
x^{2}+1})}{(x-1)+\sqrt{x^{2}+1}}\cdot \frac{(x-1)+\sqrt{x^{2}+1}}{x^{2}}%
\cdot \frac{x^{2}}{(\cos x)-1}\cdot \frac{(\cos x)-1}{\ln (1+\cos x-1)}
$$
$$
\lim_{x\rightarrow 0^{+}}\frac{\ln (1+(x-1)+\sqrt{x^{2}+1})}{(x-1)+\sqrt{%
x^{2}+1}}=\lim_{y\rightarrow 0}\frac{\ln (1+y)}{y}=1,\ \ \ \ \ \ y=(x-1)+%
\sqrt{x^{2}+1}
$$
$$
\lim_{x\rightarrow 0^{+}}\frac{(x-1)+\sqrt{x^{2}+1}}{x^{2}}\overset{H-Rule}{=%
}\lim_{x\rightarrow 0^{+}}\frac{1+\frac{x}{\sqrt{x^{2}+1}}}{2x}=\frac{1}{%
0^{+}}=+\infty 
$$
$$
\lim_{x\rightarrow 0^{+}}\frac{x^{2}}{(\cos x)-1}=-2.
$$
$$
\lim_{x\rightarrow 0^{+}}\frac{(\cos x)-1}{\ln (1+\cos x-1)}%
=\lim_{z\rightarrow 0}\frac{z}{\ln (1+z)}=1,\ \ \ \ \ \ z=(\cos x)-1
$$
Therefore
$$
\lim_{x\rightarrow 0^{+}}\frac{\ln (x+\sqrt{x^{2}+1})}{\ln {(\cos {x})}}%
=1\cdot (+\infty )\cdot (-2)\cdot 1=-\infty 
$$
EDIT. It is possible to compute the second limit without using L'Hospital's
rule. Indeed,
$$
\begin{eqnarray*}
\lim_{x\rightarrow 0^{+}}\frac{(x-1)+\sqrt{x^{2}+1}}{x^{2}}
&=&\lim_{x\rightarrow 0^{+}}\frac{\left( (x-1)+\sqrt{x^{2}+1}\right) \left(
(x-1)-\sqrt{x^{2}+1}\right) }{x^{2}\left( (x-1)-\sqrt{x^{2}+1}\right) } \\
&=&\lim_{x\rightarrow 0^{+}}\frac{\left( (x-1)^{2}-(x^{2}+1)\right) }{%
x^{2}\left( (x-1)-\sqrt{x^{2}+1}\right) }=\lim_{x\rightarrow 0^{+}}\frac{-2x%
}{x^{2}\left( (x-1)-\sqrt{x^{2}+1}\right) } \\
&=&\lim_{x\rightarrow 0^{+}}\frac{-2}{x\left( (x-1)-\sqrt{x^{2}+1}\right) }=%
\frac{-2}{0^{+}\cdot (-2)}=\frac{1}{0^{+}}=+\infty .
\end{eqnarray*}
$$
