Value of $L +\frac{153}{L}=$ 
If $\displaystyle L = \lim_{x\rightarrow 0}\bigg(\frac{1}{\ln(1+x)}-\frac{1}{\ln(x+\sqrt{x^2+1})}\bigg).$
Then value of $\displaystyle L +\frac{153}{L}=$

what i try
$$L=\lim_{x\rightarrow 0}\frac{\ln(x+\sqrt{x^2+1})-\ln(1+x)}{\ln(1+x)\ln(x+\sqrt{x^2+1})}$$
from D L Hopital rule
$$L=\lim_{x\rightarrow 0}\frac{(x^2+1)^{-\frac{1}{2}}-(1+x)^{-1}}{\ln(x+\sqrt{x^2+1})(1+x)^{-1}+\ln(1+x)(x^2+1)^{-\frac{1}{2}}}$$
How do i solve it help me please
 A: We have that
$$\ln(x+\sqrt{x^2+1})=\ln (\sqrt{x^2+1})+\ln\left(1+\frac x{\sqrt{1+x^2}}\right)=$$
$$=\frac12\ln (1+x^2)+\ln\left(1+x\left(1-\frac12x^2+O(x^4)\right)\right)=$$
$$=\frac12\ln (1+x^2)+\ln\left(1+x-\frac12x^3+O(x^5)\right)=$$
$$=\frac12x^2+x-\frac12x^2+O(x^3)=x+O(x^3)$$
then
$$\frac1{\ln(x+\sqrt{x^2+1})}=\frac1x+O(x)$$
and
$$\log(1+x)=x-\frac12 x^2+O(x^3) \implies \frac1{\ln(1+x)}=\frac1x+\frac12+O(x)$$
and
$$\frac{1}{\ln(1+x)}-\frac{1}{\ln(x+\sqrt{x^2+1})}=\frac12+O(x) \to \frac12$$
A: $$L=\lim_{x\rightarrow 0}\frac{\ln(x+\sqrt{x^2+1})-\ln(1+x)}{\ln(1+x)\ln(x+\sqrt{x^2+1})}=\\
\lim_{x\rightarrow 0}\frac{\ln\left(1+\left(\frac{x+\sqrt{x^2+1}}{1+x}-1\right)\right)}{x\frac{\ln(1+x)}{x}\cdot\frac{\ln(x+\sqrt{x^2+1})}{x+\sqrt{x^2+1}-1}\cdot\left(x+\sqrt{x^2+1}-1\right)}=\\
\lim_{x\rightarrow 0}\frac{\frac{\ln\left(1+\left(\frac{x+\sqrt{x^2+1}}{1+x}-1\right)\right)}{\frac{x+\sqrt{x^2+1}}{1+x}-1}\left(\frac{x+\sqrt{x^2+1}}{1+x}-1\right)}{x\left(x+\sqrt{x^2+1}-1\right)}=\\
\lim_{x\rightarrow 0}\frac{\frac{x+\sqrt{x^2+1}}{1+x}-1}{x\left(x+\sqrt{x^2+1}-1\right)}=\\
\lim_{x\rightarrow 0}\frac{x+\sqrt{x^2+1}-x-1}{x\left(x+\sqrt{x^2+1}-1\right)(1+x)}=\\
\lim_{x\rightarrow 0}\frac{\frac{\left(\sqrt{x^2+1}-1\right)\left(\sqrt{x^2+1}+1\right)}{\sqrt{x^2+1}+1}}{x\left(x+\frac{\left(\sqrt{x^2+1}-1\right)\left(\sqrt{x^2+1}+1\right)}{\sqrt{x^2+1}+1}\right)(1+x)}=\\
\lim_{x\rightarrow 0}\frac{\frac{x^2}{\sqrt{x^2+1}+1}}{x\left(x+\frac{x^2}{\sqrt{x^2+1}+1}\right)(1+x)}=\\
\lim_{x\rightarrow 0}\frac{\frac{1}{\sqrt{x^2+1}+1}}{\left(1+\frac{x}{\sqrt{x^2+1}+1}\right)(1+x)}=\frac12$$
