Question

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

Answer 1

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

On rationalising numerator, we get

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

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

$=\underset{x\to \infty }{lim}\frac{cx}{(\sqrt{x^2+cx}+x)}$

$=\underset{x\to \infty }{lim}\frac{cx}{(x\sqrt{1+\frac{c}{x}}+x)}$

$=\underset{x\to \infty }{lim}\frac{c}{(\sqrt{1+\frac{c}{x}}+1)}$

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

$=\underset{x\to \infty }{lim}\frac{c}{(\sqrt{1+c\left(0\right)}+1)}$

$=\frac{c}{2}$

Therefore,
$\underset{x\to \infty }{lim}\sqrt{x^2+cx}-x=\frac{c}{2}$