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

L'Hospital's rule:
$$\lim_{x \to \infty} x \{\ln(1+\sqrt{x^2 + 1}) - \ln x \} =\lim_{x \to \infty} \frac{(\ln(1+\sqrt{x^2 + 1}) - \ln x)' }{(\frac{1}{x})'} =$$$$= \lim_{x \to \infty} \frac{\frac{1}{1+\sqrt{x^2 + 1}}\cdot\frac{x}{\sqrt{x^2 + 1}} - \frac{1} {x} }{-\frac{1}{x^2}} = \lim_{x \to \infty}(x-\frac{x^2}{1+\sqrt{x^2 + 1}}\cdot\frac{x}{\sqrt{x^2 + 1}})=$$$$=\lim_{x \to \infty}\frac{1}{1+\sqrt{x^2 + 1}}\cdot\frac{x}{\sqrt{x^2 + 1}}[(1+\sqrt{x^2 + 1})\cdot\sqrt{x^2+1}-x^2]=$$$$=\lim_{x \to \infty}\frac{1}{1+\sqrt{x^2 + 1}}\cdot\frac{x}{\sqrt{x^2 + 1}}(\sqrt{x^2 + 1}+x^2+1-x^2)=\lim_{x \to \infty}\frac{x}{\sqrt{x^2 + 1}}=1.$$
