Question

simplify 1-cosx/(x^2)

It's a limit as x approaches zero question...

Answer 1

I get to this point and don't know where to go:
1-cosx/1 * 1/x^2

Answer 2

Help, please?

Answer 3

That honestly can't be simplified any further. The only thing that I can think of is the small angle approximation \(\cos x \approx 1-x^2\) so\[1- \frac{\cos x}{x^2} \approx 1- \frac{1-x^2}{x^2} = 2-\frac{1}{x^2}\]But, again, that is only an approximation.

Answer 4

Okay, thanks anyways! They say the answer is 1/2, but I have no idea how to get there...

Answer 5

Are you sure that you wrote the problem down correctly?

Answer 6

It's another limit as x approaches 0 question, if that helps...

Answer 7

Well, that would be very different than just to simplify that expression. You must have meant\[\lim_{x \rightarrow 0} \frac{1-\cos x}{x^2},\]yes?

Answer 8

Yes!! I was trying to simplify it to see if I could use any tricks, but I failed there...

Answer 9

No, but the small angle approximation can help you. The small angle approximation holds for \(x\approx 0\), so it's okay in the limit \(x\rightarrow 0\). It's actually exact in the context of the limit, and not just an approximation.\[\lim_{x\rightarrow 0} \frac{1-(1-x^2)}{x^2}=\lim_{x\rightarrow 0} 1 = \boxed{1}\]

Answer 10

Okay, that makes sense, but the answer key says 1/2...

Answer 11

Whoops, that's because I made an error. Use\[\cos x \approx 1 - \frac{x^2}{2}.\]Same concept applies, though. If you know L'Hôpital's rule, I'd just use that.

Answer 12

L'what's rule?! Agh... math+me= x_x Thanks sooo much for all the help!!