## Calculus 10th Edition

$\lim\limits_{x\to0}\dfrac{\sqrt{4+x}-2}{x}=\dfrac{1}{4}.$
$f(x)=\dfrac{\sqrt{4+x}-2}{x}=\dfrac{\sqrt{4+x}-2}{x}\times\dfrac{\sqrt{4+x}+2}{\sqrt{4+x}+2}$ $=\dfrac{(\sqrt{4+x})^2-(2)^2}{x(\sqrt{4+x}+2)}=\dfrac{x}{x(\sqrt{4+x}+2)}=\dfrac{1}{\sqrt{4+x}+2}=g(x).$ 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 $g(x)$. $\lim\limits_{x\to0}\dfrac{\sqrt{4+x}-2}{x}=\lim\limits_{x\to0}\dfrac{1}{\sqrt{4+x}+2}=\dfrac{1}{\sqrt{4+0}+2}=\dfrac{1}{4}.$