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

Feb 16, 2017

This is not an indeterminate form. See below.

#### Explanation:

As $x \rightarrow 0$, the numerator goes to $2$. Therefore, the form is not indeterminate and it would be an error to use l'Hospital's Rule.

As $x \rightarrow 0$, the denominator goes to $0$, so the limit from the right is either $\infty$ or $- \infty$.

To determine which, consider an $x$, just a bit greater than $0$.

Then $x + 1$ is a bit more than $1$, and $\sqrt{x + 1}$ is a bit greater than $1$. So the denominator is a positive number near $0$.

We can write the form of the limit as $\frac{2}{0} ^ +$ which tells us that, as $x \rightarrow {0}^{+}$, the quotient increases without bound.

We write

${\lim}_{x \rightarrow {0}^{+}} \frac{\sqrt{x + 1} + 1}{\sqrt{x + 1} - 1} = \infty$

Feb 16, 2017

There is no limit.

#### Explanation:

Generally:

${\lim}_{x \to 0} \frac{\sqrt{x + 1} + 1}{\sqrt{x + 1} - 1}$

$= {\lim}_{x \to 0} \frac{\left(1 + \frac{1}{2} x + \ldots\right) + 1}{\left(1 + \frac{1}{2} x + \ldots\right) - 1}$

$= {\lim}_{x \to 0} \frac{\left(2 + \frac{1}{2} x + \ldots\right)}{\left(\frac{1}{2} x + \ldots\right)}$

For small positive $x$, this is always going to be positive. For small negative $x$, this is always going to be negative.

So there is no limit :(

We have a 2-sided limit.