Question

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

Answer 1

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

It is of $(\infty -\infty )$ form.

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

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

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

$=\underset{x\to \infty }{lim}\frac{7x}{(x\sqrt{1+\frac{7}{x}}+x)}$

$=\underset{x\to \infty }{lim}\frac{7x}{(x\sqrt{1+\frac{7}{x}}+1)}$

$=\underset{x\to \infty }{lim}\frac{7}{(\sqrt{1+\frac{7}{x}}+1)}$

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

$=\frac{7}{(\sqrt{1+7\left(0\right)}+1)}=\frac{7}{(1+1)}=\frac{7}{2}$

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