Question

${lim}_{x\to -\infty }(\sqrt{x^2-8x}+x)$

Answer 1

${lim}_{x\to -\infty }(\sqrt{x^2-8x}+x)$

=${lim}_{y\to \infty }(\sqrt{y^2+8y}-y)$,where y=-x

$asx\to \infty ,\frac{1}{x}\to 0$

on rationalising

$={lim}_{y\to \infty }\frac{(\sqrt{y^2+8y}-y)(\sqrt{y^2+8y}+y)}{(\sqrt{y^2+8y}+y)}$

$={lim}_{y\to \infty }\frac{y^2+8y-y}{(\sqrt{y^2+8y}+y)}$

$={lim}_{y\to \infty }\frac{8y}{y\sqrt{1+\frac{8}{y}+1}}\left[\frac{\infty }{\infty }form\right]$

$={lim}_{y\to \infty }\frac{8}{\sqrt{1+\frac{8}{y}+1}}=\frac{8}{1+1}=\frac{8}{2}$

=4