(Probably) Easy Limits question

So the question is, what is the limit for when x approaches zero, for the equation:


[1-cos(x)]/x

Anybody see how to solve? Been working on it for ages now

or the limit of 1/x- cos(x)/x. The first one tends to infinite, while the latter is the fundamental limit when x->0, which is 1. Overall, I think the equation tends to infinite.

Think about that. Do you really believe that $\frac{\cos(x)}{x} \to 1$ as $x \to 0$?

I know it!

If you look closely, there's a 0/0 indeterminate form. You can now use L'Hospital Rule, taking the derivate of both the numerator and the denominator separately. (1-cos(x))' = sin(x); x'=1. The limit becomes lim x->0 of sin(x) which is 0.

(I'm still embarrassed that I supposed lim cos(x)/x -> 0....).

That's right, (although I wish you hadn't posted it, it's essentially a full solution and we're trying to help the OP) but it's a far simpler argument to simply use the definition of a derivative, as I was alluding to in the first post.

Indeed, my mistake. This applies only to sin(x), arcsin(x), tan(x), arctan(x) Thanks for letting me know! :P

I know it!

If you look closely, there's a 0/0 indeterminate form. You can now use L'Hospital Rule, taking the derivate of both the numerator and the denominator separately. (1-cos(x))' = sin(x); x'=1. The limit becomes lim x->0 of sin(x) which is 0.

(I'm still embarrassed that I supposed lim cos(x)/x -> 0....).

Thank you both very much, will think to use l'Hopitals rule next time I can't figure this type of thing out.