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