# Limits of '0/0' functions

• March 12th 2010, 10:20 AM
Limits of '0/0' functions
How can find a limit, or prove it doesn't exist, for the following function:

$\lim_{x\rightarrow 0}{\frac{sin(x)}{\sqrt{1-cos(x)}}}$

When I differentiate, I get another function with a similar problem, and it just gets more complicated, so I can assume l'Hôpital is not what I need to use here.

I tried to simplify the function by multiplying it by $\frac{\sqrt{1+cos(x)}}{\sqrt{1+cos(x)}}$ , and then I got the function: $\sqrt{1+cos(x)}$ but the changed the function and I guess I can't use it (moreover, it shows the limits are $\sqrt{2}$ and $-\sqrt{2}$, and in the original function I found that it doesn't go straight to this point, but 'gets crazy' and reaches infinity when it gets closer to 0 (from both sides).
• March 12th 2010, 10:25 AM
Krizalid
there's no limit because the one sided ones are different, since $\underset{x\to 0}{\mathop{\lim }}\,\frac{\sin x}{\sqrt{1-\cos x}}=\underset{x\to 0}{\mathop{\lim }}\,\frac{\sqrt{1+\cos x}\sin x}{\left| \sin x \right|}.$
• March 12th 2010, 11:25 AM