Question

Evaluate the limit:
$\underset{x\to -\infty }{lim}(\sqrt{4x^2-7x}+2x)$

Answer 1

We have,
$\underset{x\to -\infty }{lim}(\sqrt{4x^2-7x}+2x)$

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

$=\underset{y\to \infty }{lim}(\sqrt{4(-y{)}^2-7(-y)}+2(-y\left)\right)$

$=\underset{y\to \infty }{lim}(\sqrt{4y^2+7y}-2y)$

On rationalising numerator, we get
$=\underset{y\to \infty }{lim}\frac{(\sqrt{4y^2+7y}-2y)(\sqrt{4y^2+7y}+2y)}{(\sqrt{4y^2+7y}+2y)}$

$=\underset{y\to \infty }{lim}\frac{\left(\right(\sqrt{4y^2+7y}{)}^2-\left(2y{)}^2\right)}{(\sqrt{4y^2+7y}+2y)}$
$[\because (a-b\left)\right(a+b)=a^2-b^2]$

$=\underset{y\to \infty }{lim}\frac{(4y^2+7y-4y^2)}{(\sqrt{4y^2+7y}+2y)}$

$=\underset{y\to \infty }{lim}\frac{7y}{(\sqrt{4y^2+7y}+2y)}$

$=\underset{y\to \infty }{lim}\frac{7y}{y(\sqrt{4+\frac{7}{y}}+2)}$

$=\underset{y\to \infty }{lim}\frac{7}{\sqrt{4+\frac{7}{y}}+2}$

When $y\to \infty$, then $\frac{1}{y}\to 0$

$=\frac{7}{\left(\sqrt{4+7\left(0\right)+2}\right)}=\frac{7}{(\sqrt{4}+2)}=\frac{7}{4}$

Therefore,
$\underset{x\to -\infty }{lim}(\sqrt{4x^2-7x}+2x)=\frac{7}{4}$