Limit of Radical Expressions

In this tutorial we shall drive of limit of a radical expression, these types of limits usually solve 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, so 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}

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