# What is the limit of (2x^2 + 1)^0.5 / (3x -5) as x goes to infinity?

Apr 23, 2016

$\frac{\sqrt{2}}{3}$.

#### Explanation:

As $x \to \infty , \frac{1}{x} \mathmr{and} \frac{1}{x} ^ 2 \to 0$.

The given expression is $\frac{{\left({x}^{2} \left(2 + \frac{1}{x} ^ 2\right)\right)}^{0.5}}{x \left(3 - \frac{5}{x}\right)}$
=(((x^2)^0.5)((2+1/x^2)^0.5))/((x(3-5/x))
=(x(2+1/x^2)^0.5)/(x(3-5/x)
$= \frac{{\left(2 + \frac{1}{x} ^ 2\right)}^{0.5}}{3 - \frac{5}{x}}$.

As, $x \to \infty , \frac{1}{x} ^ 2 \mathmr{and} \frac{1}{x} \to 0 \mathmr{and}$ the given expression $\to {2}^{0.5} / 3 = \frac{\sqrt{2}}{3}$