Question

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

Answer 1

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

$=\underset{x\to \infty }{lim}\frac{x}{x\sqrt{4+\frac{1}{x}}}$

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