Evaluating Limits Involving Radicals

In this tutorial we shall discuss an example to evaluating limits involving radicals expression, in most of the cases if limit involves radical signs we shall use the method for limits is rationalization.

Let us consider an example which involve quadratic expression

\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {1 + x} - \sqrt {1 - x} }}{x}

We observe that if we directly apply directly limit in the denominator of the given function, then the result will be undefined and limit will does not exist, so first we shall rationalize the expression involving radical signs , we have

Now

\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 the x cancel each other which cause 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