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    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
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    (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 - \frac{\cos(x) - \cos(0)}{x-0} . Recognize that from anywhere?
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    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.
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    (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  \frac{\cos(x)}{x} \to 1 as  x \to 0 ?
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    (Original post by IrrationalNumber)
    Think about that. Do you really believe that  \frac{\cos(x)}{x} \to 1 as  x \to 0 ?
    Indeed, my mistake. This applies only to sin(x), arcsin(x), tan(x), arctan(x) Thanks for letting me know! :P
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    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....).
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    (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.
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    (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.
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    (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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