# (Probably) Easy Limits question watch

1. 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
2. (Original post by beckaroo7)
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
It's the limit of . Recognize that from anywhere?
3. 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.
4. (Original post by Subbi)
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 as ?
5. (Original post by IrrationalNumber)
Think about that. Do you really believe that as ?
Indeed, my mistake. This applies only to sin(x), arcsin(x), tan(x), arctan(x) Thanks for letting me know! :P
6. 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....).
7. (Original post by Subbi)
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.
8. (Original post by IrrationalNumber)
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.
(Original post by Subbi)
Indeed, my mistake. This applies only to sin(x), arcsin(x), tan(x), arctan(x) Thanks for letting me know! :P
Thank you both very much, will think to use l'Hopitals rule next time I can't figure this type of thing out.
9. (Original post by Subbi)
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....).

(Original post by beckaroo7)
Thank you both very much, will think to use l'Hopitals rule next time I can't figure this type of thing out.
Be careful about using l'Hopital's rule for this.

As IrrationalNumber pointed out early on, the ratio you're working out is -(cos x - cos 0)/(x - 0) as x->0 which is the definition of -1 times the derivative of cos x at x = 0! So if you use l'Hopital you're basically using a circular argument!!

If you're doing an applied course, you may be able to get away with this, but it would be a problem in a pure maths course, so check with your lecturer!

This is one one those questions where it helps to know what you're allowed to assume. For example, if you define cos x by its infinite power series then it's pretty clear what (1 - cos x)/x looks like as a power series too, and hence what happens as x approaches 0.

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