# Trigonometric Limit

• Apr 7th 2008, 09:57 PM
Flay
Trigonometric Limit
Can anyone help me out with this question;

Find the limit (as x approaches 0) of $\displaystyle \frac{1 - \cos2x}{x^2}$.
• Apr 7th 2008, 10:53 PM
angel.white
Quote:

Originally Posted by Flay
Can anyone help me out with this question;

Find the limit (as x approaches 0) of $\displaystyle \frac{1 - \cos2x}{x^2}$.

Using the identity $\displaystyle cos(2x) = cos^2(x) - sin^2(x)$

$\displaystyle \lim_{x\to 0} \frac{1-cos(2x)}{x^2}$

$\displaystyle =\lim_{x\to 0} \frac{1-cos^2(x)+sin^2(x)}{x^2}$

$\displaystyle =\lim_{x\to 0} \frac{sin^2(x)+sin^2(x)}{x^2}$

$\displaystyle =2\lim_{x\to 0} \frac{sin^2(x)}{x^2}$

$\displaystyle =2 \left( \lim_{x\to 0}\frac{sin(x)}x\right)^2$

Now $\displaystyle \lim_{x\to 0}\frac{sin(x)}x$ Is well known to equal one. I was going to provide a proof from my book, but you really need to see the illustrations for it to make sense. Just know that it is proven with the squeeze theorem (and should be in your book as well, check where they talk about the squeeze theorem, and if you don't see it there, check where they start talking about derivatives of trigonometric functions. Then you can just say "according to theorem ### on page ###, $\displaystyle \lim_{x\to 0}\frac{sin(x)}x=1$"

So, we apply the limit and get 1 for it and this becomes

$\displaystyle =2 \left( 1\right)^2$

$\displaystyle =2$
• Apr 8th 2008, 10:08 AM
Krizalid
Quote:

Originally Posted by angel.white
Now $\displaystyle \lim_{x\to 0}\frac{sin(x)}x$ Is well known to equal zero.

Of course, you meant to say $\displaystyle 1.$
• Apr 8th 2008, 10:28 AM
angel.white
Quote:

Originally Posted by Krizalid
Of course, you meant to say $\displaystyle 1.$

Thank you, fixed it.