Question

Compute the limit of the following as x approaches infinity.

Answer 1

Yes, that's the correct limit.

$\sqrt{x^2+1}-x=\dfrac1{\sqrt{x^2+1}+x}=\dfrac1{\sqrt{x^2}\sqrt{1+\frac1{x^2}}+x}$

$\sqrt{x^2}=|x|$, but since $x\to\infty$, we are considering $x>0$, for which $|x|=x$. Then

$\displaystyle\lim_{x\to\infty}\sqrt{x^2+1}-x=\lim_{x\to\infty}\frac1{x\left(\sqrt{1+\frac1{x^2}}+1\right)}$

and both $\dfrac1x$ and $\dfrac1{x^2}$ vanish as $x\to\infty$, making the overall limit 0.