# How do you find lim (4x)/sqrt(2x^2+1) as x->oo using l'Hospital's Rule or otherwise?

Dec 14, 2016

I would rewrite the quotient.

#### Explanation:

lim_(xrarroo)(4x)/sqrt(2x^2+1) = lim_(xrarroo)(4x)/(xsqrt(2+1/x^2)

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

Note that

For $x \ne 0$, we have $\sqrt{2 {x}^{2} + 1} = \sqrt{{x}^{2}} \sqrt{2 + \frac{1}{x} ^ 2}$

Since $\sqrt{{x}^{2}} = \left\mid x \right\mid$, and we are interested in the limit as $x$ increases without bound, we have $\left\mid {x}^{2} \right\mid = x$

For x decreasing without bound, we get

$\sqrt{{x}^{2}} = - x$.

So the limit as $x \rightarrow - \infty$ would be $- 2 \sqrt{2}$.

Here is the graph:

graph{(4x)/(sqrt(2x^2+1) [-16.01, 16.02, -8.01, 8]}