Question

Limit   

Answer 1

Rationalize the numerator:

$\dfrac{\sqrt{x+4}-2}x\cdot\dfrac{\sqrt{x+4}+2}{\sqrt{x+4}+2}=\dfrac{(\sqrt{x+4})^2-2^2}{x(\sqrt{x+4}+2)}=\dfrac x{x(\sqrt{x+4}+2)}=\dfrac1{\sqrt{x+4}+2}$

This is continuous at $x=0$, so we can evaluate the limit directly by substitution:

$\displaystyle\lim_{x\to0}\frac{\sqrt{x+4}-2}x=\lim_{x\to0}\frac1{\sqrt{x+4}+2}=\frac1{\sqrt4+2}=\frac14$