Question

Evaluate the limit:
$\underset{x\to \infty }{lim}\frac{x}{\sqrt{4x^2+1}-1}$

Answer 1

We have,
$\underset{x\to \infty }{lim}\frac{x}{\sqrt{4x^2+1}-1}$

$=\underset{x\to \infty }{lim}\frac{x}{\sqrt{x^2(4+\frac{1}{x^2})}-1}$

$=\underset{x\to \infty }{lim}\frac{x}{x(\sqrt{(4+\frac{1}{x^2})}-\frac{1}{x})}$

$=\underset{x\to \infty }{lim}\frac{1}{(\sqrt{(4+\frac{1}{x^2})}-\frac{1}{x})}$

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

$=\frac{1}{(\sqrt{(4+0)}-0)}$

$=\frac{1}{\sqrt{4}}=\frac{1}{2}$