How do you find lim x^2/(sqrt(2x+1)-1) as x->0 using l'Hospital's Rule?

Aug 29, 2017

We start by checking to make sure the limit is actually of the form $\frac{0}{0}$ or $\frac{\infty}{\infty}$.

$L = {0}^{2} / \left(\sqrt{2 \left(0\right) + 1} - 1\right) = \frac{0}{1 - 1} = \frac{0}{0}$

So we may indeed use l'hospital's.

The derivative of $\sqrt{2 x + 1}$ is $\frac{2}{2 \sqrt{2 x + 1}} = \frac{1}{\sqrt{2 x + 1}}$.

$L = {\lim}_{x \to 0} \frac{2 x}{\frac{1}{\sqrt{2 x + 1}}}$

$L = {\lim}_{x \to 0} 2 x \sqrt{2 x + 1}$

If we evaluate now, we get:

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

Looking at the graph, we realize that $y$ does approach $0$ as $x$ approaches $0$.

graph{x^2/(sqrt(2x+ 1) - 1) [-4.93, 4.934, -2.465, 2.465]}

Hopefully this helps!