Question Number 73017 by Rio Michael last updated on 05/Nov/19
$${find}\: \\ $$$$\:\underset{{x}\rightarrow+\infty} {\mathrm{lim}}\:\:{x}\sqrt{{x}^{\mathrm{2}} \:+\:\mathrm{1}}\: \\ $$
Commented by mathmax by abdo last updated on 05/Nov/19
$$\left.={lim}_{{x}\rightarrow+\infty} {x}\sqrt{{x}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right.}\right)={lim}_{{x}\rightarrow+\infty} {x}^{\mathrm{2}} \sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }}={lim}_{{x}\rightarrow+\infty} {x}^{\mathrm{2}} =+\infty \\ $$