Answer
$\lim\limits_{x\to0}\dfrac{\sqrt{x+5}-\sqrt{5}}{x}=\dfrac{1}{2\sqrt{5}}.$
Work Step by Step
$f(x)=\dfrac{\sqrt{x+5}-\sqrt{5}}{x}=\dfrac{\sqrt{x+5}-\sqrt{5}}{x}\times\dfrac{\sqrt{x+5}+\sqrt{5}}{\sqrt{x+5}+\sqrt{5}}$
$=\dfrac{(\sqrt{x+5})^2-(\sqrt{5})^2}{x(\sqrt{x+5}+\sqrt{5})}=\dfrac{x}{x(\sqrt{x+5}+\sqrt{5})}$
$=\dfrac{1}{\sqrt{x+5}+\sqrt{5}}=g(x).$
The function $g(x)$ agrees with the function $f(x)$ at all points except $x=0$. Therefore we find the limit as x approaches $0$ of $f(x)$ by substituting the value into $g(x)$.
$\lim\limits_{x\to0}\dfrac{\sqrt{x+5}-\sqrt{5}}{x}=\lim\limits_{x\to0}\dfrac{1}{\sqrt{x+5}+\sqrt{5}}=\dfrac{1}{\sqrt{0+5}+\sqrt{5}}=\dfrac{1}{2\sqrt{5}}.$
