# How do you find the limit x^2/sqrt(2x+1)-1 as x->0?

Jan 2, 2017

As asked, at $x = 0$, the denominator is not $0$. So use substitution.

For ${\lim}_{x \rightarrow 0} {x}^{2} / \left(\sqrt{2 x + 1} - 1\right)$ see below.

#### Explanation:

For $x$ close to $0$, the numerator, ${x}^{2}$ is also close to zero.

The denominator is close to $\sqrt{2 \left(0\right) + 1} = \sqrt{1} = 1$

So the quotient is close to $\frac{0}{1}$ which is $0$.

If a different quotient was intended

If the question is missing parentheses and should be x^2/(sqrt(2x+1)-1), then use

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

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

$= \frac{{x}^{2} \left(\sqrt{2 x + 1} + 1\right)}{2 x}$

$= \frac{x \left(\sqrt{2 x + 1} + 1\right)}{2}$.

So, the limit is $\frac{0 \left(\sqrt{2 \left(0\right) + 1} + 1\right)}{2} = 0$