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
 A: L'Hospital's rule:
$$\lim_{x \to \infty} x \{\ln(1+\sqrt{x^2 + 1}) - \ln x \} =\lim_{x \to \infty} \frac{(\ln(1+\sqrt{x^2 + 1}) - \ln x)' }{(\frac{1}{x})'} =$$$$= \lim_{x \to \infty} \frac{\frac{1}{1+\sqrt{x^2 + 1}}\cdot\frac{x}{\sqrt{x^2 + 1}} - \frac{1} {x} }{-\frac{1}{x^2}} = \lim_{x \to \infty}(x-\frac{x^2}{1+\sqrt{x^2 + 1}}\cdot\frac{x}{\sqrt{x^2 + 1}})=$$$$=\lim_{x \to \infty}\frac{1}{1+\sqrt{x^2 + 1}}\cdot\frac{x}{\sqrt{x^2 + 1}}[(1+\sqrt{x^2 + 1})\cdot\sqrt{x^2+1}-x^2]=$$$$=\lim_{x \to \infty}\frac{1}{1+\sqrt{x^2 + 1}}\cdot\frac{x}{\sqrt{x^2 + 1}}(\sqrt{x^2 + 1}+x^2+1-x^2)=\lim_{x \to \infty}\frac{x}{\sqrt{x^2 + 1}}=1.$$
A: 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 $$\lim_{x \to \infty} x \{\ln(1+\sqrt{x^2 + 1}) - \ln x \} = 1 .$$
A: The problem that you are asking about is really trivial and straightforward. No l'Hospital needed whatsoever. You are asking about the limit of x ln f (x), where f (x) goes to infinity; that's trivial. 
When you ask "Shall I change its form (minimal change)": It's not a minimal change what you are proposing. You are moving a parenthesis and absolutely totally change the expression. This is an absolutely fatal mistake. A mistake that apparently everyone trying to answer your question has made as well...
Now you might read your homework question carefully and I could imagine that since what you ask is so trivial, and changing the parenthesis makes it a much more interesting problem, that maybe you copied your homework question wrongly. 
