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
 A: 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
$\lim_{x\to 0^+} \frac{\ln(x+ \sqrt{x^2+1})}{x^2}=\lim_{x\to 0^+} {\ln(x+ \sqrt{x^2+1})^{1/x^2}}=\ln(\infty)=\infty$
A: 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,
$$
\begin{eqnarray*}
\lim_{x\rightarrow 0^{+}}\frac{(x-1)+\sqrt{x^{2}+1}}{x^{2}}
&=&\lim_{x\rightarrow 0^{+}}\frac{\left( (x-1)+\sqrt{x^{2}+1}\right) \left(
(x-1)-\sqrt{x^{2}+1}\right) }{x^{2}\left( (x-1)-\sqrt{x^{2}+1}\right) } \\
&=&\lim_{x\rightarrow 0^{+}}\frac{\left( (x-1)^{2}-(x^{2}+1)\right) }{%
x^{2}\left( (x-1)-\sqrt{x^{2}+1}\right) }=\lim_{x\rightarrow 0^{+}}\frac{-2x%
}{x^{2}\left( (x-1)-\sqrt{x^{2}+1}\right) } \\
&=&\lim_{x\rightarrow 0^{+}}\frac{-2}{x\left( (x-1)-\sqrt{x^{2}+1}\right) }=%
\frac{-2}{0^{+}\cdot (-2)}=\frac{1}{0^{+}}=+\infty .
\end{eqnarray*}
$$
A: Using asymptotics for $\ x\to0$ we have:
$$\ \sqrt{1+x^2}≈1+\frac{x^2}{2}$$
$$\ \ln(1+x)≈x$$
$$\ \cos(x)≈1-\frac{x^2}{2}$$
So your limit gets:
$$\ \lim_{x\to0^+}\frac{\ln(x+\sqrt{1+x^2})}{\ln(\cos x)}≈\lim_{x\to0^+}\frac{\ln(1+x+\frac{x^2}{2})}{\ln(1-\frac{x^2}{2})}≈$$
$$≈\lim_{x\to0^+}\frac{x+\frac{x^2}{2}}{-\frac{x^2}{2}}=-\infty$$
A: Outline: Our expression, for $x\ne 0$, is equal to
$$\frac{\ln(x+\sqrt{x^2+1})}{x} \cdot\frac{x}{\ln(\cos x)}.$$ 
Note that 
$$\lim_{x\to 0} \frac{\ln(x+\sqrt{x^2+1})-0}{x}$$
is by definition the derivative of $\ln(x+\sqrt{x^2+1})$ at $x=0$. It is easy to compute that derivative. The only important thing is that it is positive.
Now we examine what happens to $\frac{x}{\ln(cos x)}$ as $x\to 0$. It is easier to look at what happens to $\frac{\ln(cosx)}{x}$. Again, think derivative.
A: Using $\lim\limits_{x\to0}\frac{\log(1+x)}x=1$ and $\lim\limits_{x\to0}\frac{\sin(x)}x=1$,
$$
\begin{align}
&\lim_{x\to0^+}\frac{\log\left(x+\sqrt{x^2+1}\right)}{\log(\cos(x))}\\
&=\lim_{x\to0^+}\frac{\log\left(1+x+\sqrt{x^2+1}-1\right)}{\log(1+\cos(x)-1)}\\
&=\lim_{x\to0^+}\frac{\log\left(1+x+\frac{x^2}{\sqrt{x^2+1}+1}\right)}{\log\left(1-\frac{\sin^2(x)}{\cos(x)+1}\right)}\\
&=\lim_{x\to0^+}\frac{x+\frac{x^2}{\sqrt{x^2+1}+1}}{\frac{\sin^2(x)}{\cos(x)+1}}
\cdot\lim_{x\to0^+}\frac{\log\left(1+x+\frac{x^2}{\sqrt{x^2+1}+1}\right)}{x+\frac{x^2}{\sqrt{x^2+1}+1}}
\cdot\lim_{x\to0^+}\frac{\frac{\sin^2(x)}{\cos(x)+1}}{\log\left(1-\frac{\sin^2(x)}{\cos(x)+1}\right)}\\
&=\lim_{x\to0^+}\frac1x
\cdot\lim_{x\to0^+}\frac{x^2}{\sin^2(x)}
\cdot\lim_{x\to0^+}\frac{1+\frac{x}{\sqrt{x^2+1}+1}}{\frac1{\cos(x)+1}}\cdot1\cdot-1\\[6pt]
&=\lim_{x\to0^+}\frac1x\cdot1\cdot2\cdot1\cdot-1\\[18pt]
&=-\infty
\end{align}
$$
A: Hint: Let $x=\sinh t$, as $t\to0^+.$ Then, since $1+\sinh^2t=\cosh^2t,$ and $\cosh t\pm\sinh t=e^{\large\pm t},~$ the numerator simply becomes t. As for the denominator, use $\cos x\simeq1-\dfrac{x^2}2~,$ in conjunction with $\ln(1+u)\simeq u$.
