## anonymous 4 years ago lim x-->0 for [(√x+4) - 2]/x its the square root of x+4 not just x

1. anonymous

This is an indeterminate form that of 0/0 thus we can apply L'Hospital's Rule to obtain the limit as 1/4.

2. anonymous

how?

3. anonymous
4. anonymous

thanks

5. anonymous

Let me know if you have any more questions :)

6. anonymous

$\lim_{x\rightarrow 0}\frac{\sqrt{x+4}-2}{x}=\lim_{x\rightarrow 0}\frac{\sqrt{x+4}-2}{x}\cdot\frac{\sqrt{x+4}+2}{\sqrt{x+4}+2}=$$=\lim_{x\rightarrow 0}\frac{x+4-4}{x(\sqrt{x+4}+2)}=\lim_{x\rightarrow 0}\frac{1}{\sqrt{x+4}+2}=\frac{1}{4}$