In this tutorial we shall discuss an example of evaluating limits involving radical expressions. In most cases, if the limit involves radical signs we shall use the method for limits known as rationalization.

Let us consider an example which involves a quadratic expression
$\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} – \sqrt {1 – x} }}{x}$

We can see that if we directly apply limit to the denominator of the given function, then the result will be undefined and this limit does not exist. So first we shall rationalize the expression involving radical signs.

We have:
$\begin{gathered} \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} – \sqrt {1 – x} }}{x} \\ = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} – \sqrt {1 – x} }}{x} \times \frac{{\sqrt {1 + x} + \sqrt {1 – x} }}{{\sqrt {1 + x} + \sqrt {1 – x} }} \\ = \mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {\sqrt {1 + x} } \right)}^2} – {{\left( {\sqrt {1 – x} } \right)}^2}}}{{x\left( {\sqrt {1 + x} + \sqrt {1 – x} } \right)}} \\ = \mathop {\lim }\limits_{x \to 0} \frac{{1 + x – \left( {1 – x} \right)}}{{x\left( {\sqrt {1 + x} + \sqrt {1 – x} } \right)}} \\ = \mathop {\lim }\limits_{x \to 0} \frac{{1 + x – 1 + x}}{{x\left( {\sqrt {1 + x} + \sqrt {1 – x} } \right)}} \\ = \mathop {\lim }\limits_{x \to 0} \frac{{2x}}{{x\left( {\sqrt {1 + x} + \sqrt {1 – x} } \right)}} \\ \end{gathered}$

Now $x$ cancel each other out, which causes the limit to be undefined
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} – \sqrt {1 – x} }}{x} = \mathop {\lim }\limits_{x \to 0} \frac{2}{{\sqrt {1 + x} + \sqrt {1 – x} }}$

Applying the limits, we have
$\Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} – \sqrt {1 – x} }}{x} = \frac{2}{{\sqrt {1 + 0} + \sqrt {1 – 0} }} = \frac{2}{{\sqrt 1 + \sqrt 1 }} = \frac{2}{2} = 1$