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# Find the limit of [1-cos(x)]/x Watch

1. Err, have you guys don L^Hopitals yet? If you plug in x = 0 you get an indeterminate form, so differentiate and then you'll get lim x -> 0 sin(x) = 0. Hence lim x-> 0 of 1-cos(x)/x = 0
2. (Original post by barnisaurusrex)
Err, have you guys don L^Hopitals yet? If you plug in x = 0 you get an indeterminate form, so differentiate and then you'll get lim x -> 0 sin(x) = 0. Hence lim x-> 0 of 1-cos(x)/x = 0
I would strongly advise not using L'Hopital's rule for this question. This is because the question is essentially asking you to calculate a derivative. The question could reasonably then state 'Hence show that the cosine function is differentiable at 0 with derivative 0.' Using L'Hopital's rule would then be circular reasoning. In addition, it doesn't follow the hint given at all and is far too strong for this kind of question. I think that the question was designed to help practice limits, rather than to pull out big machinery.
3. (Original post by IrrationalNumber)
I would strongly advise not using L'Hopital's rule for this question. This is because the question is essentially asking you to calculate a derivative. The question could reasonably then state 'Hence show that the cosine function is differentiable at 0 with derivative 0.' Using L'Hopital's rule would then be circular reasoning. In addition, it doesn't follow the hint given at all and is far too strong for this kind of question. I think that the question was designed to help practice limits, rather than to pull out big machinery.
Well I guess that's physicists mentality vs. mathematicians. L^Hopitals gets you there perfectly fine and everything you've suggested seems overly technical and flippant. Personally I'm in favour of getting to the answer the easiest possible way, but hey, I can see if it's an analysis module then the most abstract, least intuitive way is definitely the best way

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