# Thread: The Pinching Theorem; Trig Limits

1. ## The Pinching Theorem; Trig Limits

Can anyone help me evaluate this limit?

$\lim_{x\to 0} \frac{x^2}{1-\cos2x}$

I can see that the numerator approaches 0 faster and that you can say that this is equal to $(x)(\frac{x}{1-\cos2x})$ but i'm not sure where to go from there. Oh and the answer is 1/2 and i cant use L'hospital's theorem.

Thanks!

2. Hello, discobob!

Do you have to use the "Squeeze Theorem"?

$\lim_{x\to 0} \frac{x^2}{1-\cos2x}$

We have: . $\frac{x^2}{1-\cos2x} \;=\;\frac{x^2}{2\sin^2\!x}\;=\;\frac{1}{2}\left(\ frac{x}{\sin x}\right)^2$

Therefore: . $\lim_{x\to0}\,\left[\frac{1}{2}\left(\frac{x}{\sin x}\right)^2\right] \;=\; \frac{1}{2}\cdot 1^2 \;=\;\frac{1}{2}$

3. Oh, ok. I just really need to study my trig Identities. I understand how you do it. Thanks a lot Soroban!