# When do you rationalize the denominator to calculate the limit?

Mar 10, 2015

At the risk of sounding like I'm being flippant, you rationalize the denominator when you need to and it helps.

Example 1: Evaluate: ${\lim}_{x \rightarrow 9} \frac{x}{\sqrt{x} + 5}$
The limits of the numerator and denominator are: ${\lim}_{x \rightarrow 9} x = 9$ and${\lim}_{x \rightarrow 9} \left(\sqrt{x} + 5\right) = 8$. So we can find the requested limit by using the quotient property of limits. There is no need to rationalize.

${\lim}_{x \rightarrow 9} \frac{x}{\sqrt{x} + 5} = \frac{9}{8}$

Example 2: Evaluate: ${\lim}_{x \rightarrow 4} \frac{x - 4}{\sqrt{x} - 2}$
This time the limits of the numerator and denominator are ${\lim}_{x \rightarrow 4} \left(x - 4\right) = 0$ and ${\lim}_{x \rightarrow 4} \left(\sqrt{x} - 2\right) = 0$

We cannot use the quotient rule for limits, so we'll try rationalizing the denominator to see if that helps.

(Yes, really, seriously. Try something and see if it works. There is no 'cookbook'. for solving all problems.)

$\frac{x - 4}{\sqrt{x} - 2} \frac{\sqrt{x} + 2}{\sqrt{x} + 2} = \frac{\left(x - 4\right) \left(\sqrt{x} + 2\right)}{x - 4} = \frac{\sqrt{x} + 2}{1}$ (for $x \ne 4$).

The denominator of this new ratio does not go to $0$ as $x \rightarrow 4$, so we can use the quotient property of limits. It worked. We get:

${\lim}_{x \rightarrow 4} \frac{x - 4}{\sqrt{x} - 2} = {\lim}_{x \rightarrow 4} \frac{\sqrt{x} + 2}{1} = 4$ .