Find \(\lim\limits_{x\rightarrow 5}\dfrac{\sqrt{x}-\sqrt{5}}{x-5}\).
we cannot use the Limit of a Quotient property. A different strategy is necessary. We rationalize the numerator of the quotient. \[ \begin{eqnarray*} \frac{\sqrt{x}-\sqrt{5}}{x-5}&=&\frac{( \sqrt{x}-\sqrt{5}) }{(x-5) }\cdot \frac{( \sqrt{x}+\sqrt{5}) }{( \sqrt{x}+\sqrt{5}) } =\frac{x-5}{(x-5) ( \sqrt{x}+\sqrt{5}) } \underset{\underset{\color{#0066A7}{\hbox{$x\neq 5$}}}{\color{#0066A7}{\uparrow}}}{=} \frac{1}{\sqrt{x}+\sqrt{5}} \end{eqnarray*} \]
Do you see why rationalizing the numerator works? It causes the term \(x-5\) to appear in the numerator, and since \(x\neq 5\), the factor \(x-5\) can be divided out. Then
When finding a limit, remember to include “\(\lim\limits_{x\rightarrow c}\)” at each step until you let \({x\rightarrow c}\).
\[ \begin{eqnarray*} \lim_{x\rightarrow 5}\frac{\sqrt{x}-\sqrt{5}}{x-5}&=&\lim_{x\rightarrow 5}\frac{1}{\sqrt{x}+\sqrt{5}} \underset{\underset{\color{#0066A7}{\hbox{Use the Limit of a Quotient}}}{\color{#0066A7}{\uparrow}}} {=}\frac{1 }{\sqrt{5}+\sqrt{5}}=\frac{1}{2\sqrt{5}}=\frac{\sqrt{5}}{10} \end{eqnarray*} \]