Find $\lim\limits_{x\rightarrow 5}\dfrac{\sqrt{x}-\sqrt{5}}{x-5}$.

Solution The domain of $h(x)=\dfrac{\sqrt{x}-\sqrt{5}}{x-5}$ is {${ x|x\geq 0, x\ne 5}$}. Since the limit of the denominator is $$\lim\limits_{x\rightarrow 5}g(x)=\lim\limits_{x\rightarrow 5}(x-5)=0$$

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

$$\begin{eqnarray*} \lim_{x\rightarrow 5}\frac{\sqrt{x}-\sqrt{5}}{x-5}&=&\lim_{x\rightarrow 5}\frac{1}{\sqrt{x}+\sqrt{5}} \underset{\underset{\color{#0066A7}{\hbox{Use the Limit of a Quotient}}}{\color{#0066A7}{\uparrow}}} {=}\frac{1 }{\sqrt{5}+\sqrt{5}}=\frac{1}{2\sqrt{5}}=\frac{\sqrt{5}}{10} \end{eqnarray*}$$