Yes, that's the correct limit.
[tex]\sqrt{x^2+1}-x=\dfrac1{\sqrt{x^2+1}+x}=\dfrac1{\sqrt{x^2}\sqrt{1+\frac1{x^2}}+x}[/tex]
[tex]\sqrt{x^2}=|x|[/tex], but since [tex]x\to\infty[/tex], we are considering [tex]x>0[/tex], for which [tex]|x|=x[/tex]. Then
[tex]\displaystyle\lim_{x\to\infty}\sqrt{x^2+1}-x=\lim_{x\to\infty}\frac1{x\left(\sqrt{1+\frac1{x^2}}+1\right)}[/tex]
and both [tex]\dfrac1x[/tex] and [tex]\dfrac1{x^2}[/tex] vanish as [tex]x\to\infty[/tex], making the overall limit 0.
