Question

Evaluate the limit:
$\underset{x\to -\infty }{lim}(\sqrt{x^2-8x}+x)$

Answer 1

We have,
$\underset{x\to -\infty }{lim}(\sqrt{x^2-8x}+x)$

Taking $y=-x$
If $x\to -\infty \Rightarrow -x\to \infty \Rightarrow y\to \infty$

$=\underset{y\to \infty }{lim}(\sqrt{(-y{)}^2-8(-y)}+(-y\left)\right)$

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

On rationalising numerator, we get

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

$=\underset{y\to \infty }{lim}\frac{\left(\right(\sqrt{y^2+8y}{)}^2-y^2)}{(\sqrt{y^2+8y}+y)}$
$[\because (a-b\left)\right(a+b)=a^2-b^2]$

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

$=\underset{y\to \infty }{lim}\frac{8y}{(y\sqrt{1+\frac{8}{y}}+y)}$

$=\underset{y\to \infty }{lim}\frac{8y}{y(\sqrt{1+\frac{8}{y}}+1)}$

$=\underset{y\to \infty }{lim}\frac{8}{\sqrt{1+\frac{8}{y}}+1}$
When $y\to \infty$, then $\frac{1}{y}\to 0$

$=\frac{8}{(\sqrt{1+8\left(0\right)}+1)}=\frac{8}{2}=4$

Therefore,
$\underset{x\to -\infty }{lim}(\sqrt{x^2-8x}+x)=4$