Limit $\lim_{x\to\infty}(x(\log(1+\sqrt{1+x^2}-\log(x)))$ can someone give me a hint/solution for:
$$\lim_{x\to\infty}\left(x(\log(1+\sqrt{1+x^2}-\log(x))\right)$$
Shall I do a derivative ?
But there's no L'Hospital to use here..
Shall I change its form ?
$x(\log(1+(1+x^2)^{1/2})-\log(x))$ (minimal change)
Or what shall I do ?
Thanks

for $x$ large 
$\begin{align}
\ln(1+\sqrt{x^2 + 1}) - \ln x &= \ln \left( 1+ x(1+1/x^2)^{1/2} \right) - \ln x\\
 &=\ln[ 1 + x(1 + \frac{1}{2x^2} + \cdots  )] - \ln x\\
 &= \ln[x(1 + \frac{1}{x} + \frac{1}{2x^2}+\cdots)] -\ln x\\
 &= \ln(1 + \frac{1}{x} + \frac{1}{2x^2}+\cdots) \\
 &=  \frac{1}{x} +\cdots
\end{align}$
therefore $$\lim_{x \to \infty} x \{\ln(1+\sqrt{x^2 + 1}) - \ln x \} = 1 .$$
