$\lim _{x\to \infty }x\log\left(1+\sqrt{1+x^2}-x\right)$

Question

Is there any trick to solve this?

$\lim _{x\to \infty }x\log\left(1+\sqrt{1+x^2}-\log\left(x\right)\right)$

Answer 1

Hint: Observe \begin{align} \lim_{x\rightarrow \infty} (1+\sqrt{1+x^2}-\log x )= \infty \end{align}

Answer 2

I am going to assume that the limit in question is actually

$$\lim_{x\to \infty}\,x\left(\log\left(1+\sqrt{1+x^2}\right)-\log(x))\right)$$

Proceeding accordingly, we have

$$\begin{align} \log\left(1+\sqrt{1+x^2}\right)-\log(x)&=\log\left(\frac{1+\sqrt{1+x^2}}{x}\right)\\\\ &=\log\left(\sqrt{1+\frac1{x^2}}+\frac1x \right)\\\\ &=\log\left(1+\frac1x +O\left(\frac1{x^2}\right)\right)\\\\ &=\frac1x+O\left(\frac1{x^2}\right) \end{align}$$

Therefore, we have

$$\lim_{x\to \infty}\,x\left(\log\left(1+\sqrt{1+x^2}\right)-\log(x))\right)=\lim_{x\to \infty}\left(1+O\left(\frac1x\right)\right)=1$$

Answer 3

HINT: observe that $$\log(1+\sqrt{1+x^2})-\log(x)=0$$ for $x$ tends to infinity.