Limit of Radical Expressions

In this tutorial we shall derive the limit of a radical expression. These types of limits are usually solved by using method rationalization.

Now we consider an example to evaluate:
\[\mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {k + x} – \sqrt {k – x} }}{x}\]

Since the denominator of the given expression becomes zero at $$x = 0$$, we must rationalize it before applying limits:
\[\begin{gathered} \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {k + x} – \sqrt {k – x} }}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {k + x} – \sqrt {k – x} }}{x} \times \frac{{\sqrt {k + x} + \sqrt {k – x} }}{{\sqrt {k + x} + \sqrt {k – x} }} \\ \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {k + x} – \sqrt {k – x} }}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt {k + x} – \sqrt {k – x} } \right)\left( {\sqrt {k + x} + \sqrt {k – x} } \right)}}{{x\left( {\sqrt {k + x} + \sqrt {k – x} } \right)}} \\ \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {k + x} – \sqrt {k – x} }}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{{{\left( {\sqrt {k + x} } \right)}^2} – {{\left( {\sqrt {k – x} } \right)}^2}}}{{x\left( {\sqrt {k + x} + \sqrt {k – x} } \right)}} \\ \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {k + x} – \sqrt {k – x} }}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{k + x – k + x}}{{x\left( {\sqrt {k + x} + \sqrt {k – x} } \right)}} \\ \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {k + x} – \sqrt {k – x} }}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{2x}}{{x\left( {\sqrt {k + x} + \sqrt {k – x} } \right)}} \\ \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {k + x} – \sqrt {k – x} }}{x} = \mathop {\lim }\limits_{x \to 0} \frac{2}{{\left( {\sqrt {k + x} + \sqrt {k – x} } \right)}} \\ \end{gathered} \]

Now applying the limits, we have
\[\begin{gathered} \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {k + x} – \sqrt {k – x} }}{x} = \frac{2}{{\left( {\sqrt {k + 0} + \sqrt {k – 0} } \right)}} \\ \Rightarrow \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt {k + x} – \sqrt {k – x} }}{x} = \frac{2}{{\sqrt k + \sqrt k }} = \frac{2}{{2\sqrt k }} = \frac{1}{{\sqrt k }} \\ \end{gathered} \]