Can you help me calculate a limit? Is there a way to accurately calculate
$$
\lim_{x \to 0} \frac{\sqrt{3 + x} - \sqrt{3 - x}}{\sqrt{4+x^2} - \sqrt{4-2x}}
$$
using brute force with a graphing calculator? I can't use L'Hôpital's rule and based on Wolfram Alpha's calculations, I'm doing it all wrong. It looks like I can't get an accurate form without it, so I just need a decimal. I would really appreciate it if there was some kind of explanation. I don't know anyone to go to that understands calculus, and I really want to understand it.

Hint: Multiply top and bottom by $\sqrt{3+x}+\sqrt{3-x}$, also multiply top and bottom by $\sqrt{4+x^2}+\sqrt{4-2x}$. 
There will be some easy to handle stuff, and also $\dfrac{2x}{x^2+2x}$, which has limit $1$. The final answer should be $\dfrac{2}{\sqrt{3}}$. 
Remark: Whenever $\sqrt{a}-\sqrt{b}$ gets into trouble, its friend $\sqrt{a}+\sqrt{b}$ is ready to pitch in and help. Note that $(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})=a-b$. 
