<s>[INST] 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
 [/INST] HINT:
To get rid of the $ln$ in the denominator, remember that $\ln \cos x= \frac{1}{2} \ln(1-\sin^2(x))$.
As for the $ln$ in the nominator, you'll have to calculate the limit: $\lim_{x\to 0^+} {\ln(x+ \sqrt{x^2+1})^{1/x^2}}$ wich is easier (it's zero).
SOLUTION:
\begin{align}
\lim_{x\to 0^+} \frac{\ln(x+ \sqrt{x^2+1})}{\ln{(\cos{x})}}
&=2\lim_{x\to 0^+} \frac{\ln(x+ \sqrt{x^2+1})}{\ln(1-\sin^2(x))}\tag{1}\\
&=2\lim_{x\to 0^+} \frac{\sin^2(x)}{\ln(1-\sin^2(x))} \cdot \frac{x^2}{\sin^2(x)} \cdot \frac{\ln(x+ \sqrt{x^2+1})}{x^2}\tag{2}\\
&=2\lim_{x\to 0^+} \frac{\sin^2(x)}{\ln(1-\sin^2(x))} \cdot \lim_{x\to 0^+} \frac{x^2}{\sin^2(x)} \cdot \lim_{x\to 0^+} \frac{\ln(x+ \sqrt{x^2+1})}{x^2}\tag{3}\\
&=2\cdot (-1) \cdot 1 \cdot \infty \tag{4}\\
&=-\infty \tag{5}\\
\end{align}
Explanation:  
$(1)$: $\frac{1}{2} \ln(1-\sin^2(x))=\frac{1}{2} \ln(\cos^2(x))=\frac{1}{2} \cdot 2\ln(\cos x)=\ln \cos x$
$(4)$: $\lim_{x\to 0^+} \frac{\sin^2(x)}{\ln(1-\sin^2(x))} =\lim_{x\to 0^+} \frac{1}{\ln(1-\sin^2(x))^{1/\sin^2(x)}}=\lim_{y\to 0^+} \frac{1}{\ln(1-y)^{1/y}}=\frac{1}{\ln(\frac{1}{e})}=-1 $
and
$\lim_{x\to 0^+} \frac{\ln(x+ \sqrt{x^2+1})}{x^2}=\lim_{x\to 0^+} {\ln(x+ \sqrt{x^2+1})^{1/x^2}}=\ln(\infty)=\infty$
</s>