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Math Help - Calculating a Limit

  1. #1
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    Calculating a Limit

    lim x --> 0 ((1-cos x)/ x^2)

    How would I go about calculating this limit?

    Thankyou
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  2. #2
    MHF Contributor Unknown008's Avatar
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    Maybe you can use the l'Hopital's rule twice.

    That is, you differentiate the numerator and denominator with respect to x twice.
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  3. #3
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    Unfortunately, we're not yet allowed to use that rule. At this stage we're only supposed to be assuming that lim x--> sin x/ x = 1 and lim x--> 0 (1- cos x)/ x = 0. Thankyou for the input anyway though!
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  4. #4
    MHF Contributor harish21's Avatar
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    Quote Originally Posted by katedew987 View Post
    Unfortunately, we're not yet allowed to use that rule. At this stage we're only supposed to be assuming that lim x--> sin x/ x = 1 and lim x--> 0 (1- cos x)/ x = 0. Thankyou for the input anyway though!
    In your first post you have written that you want to find lim x --> 0 ((1-cos x)/ x^2)

    and in your second post,you said you wanted to find lim x --> 0 ((1-cos x)/ x)

    the former one \neq 0
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  5. #5
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    No, sorry, perhaps I didn't make it clear enough, I'm supposed to be finding lim x--> 0 ((1-cosx)/ x^2) working on the assumption that lim x--> (1 - cos x) = 0
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  6. #6
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    How did you do the original proof that
    \lim\limits_{x\to0} \frac{1-cos(x)}{x}=0 ?
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  7. #7
    MHF Contributor Also sprach Zarathustra's Avatar
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    Use 2sin^2(\frac{x}{2})=1-cos(x)

    And use:

    YOU will get that the limit is 1/2
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  8. #8
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by katedew987 View Post
    lim x --> 0 ((1-cos x)/ x^2)

    How would I go about calculating this limit?

    Thankyou
    \displaystyle \lim_{x\to 0}\frac{1-\cos(x)}{x^2}=\lim_{x\to 0}\left(\frac{\sin(x)}{x}\right)^2\cdot\frac{1}{1+  \cos(x)}
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  9. #9
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by katedew987 View Post
    lim x --> 0 ((1-cos x)/ x^2)

    How would I go about calculating this limit?

    Thankyou
    Starting from the 'fundamental limit'...

    \displaystyle \lim_{x \rightarrow 0} \frac{\sin x}{x}= 1 (1)

    ... You first derive...

    \displaystyle \lim_{x \rightarrow 0} \frac{\sin^{2} x}{x^{2}} = \lim_{x \rightarrow 0} \frac{1-\cos^{2} x}{x^{2}} = \lim_{x \rightarrow 0} \frac{(1+\cos x)\ (1-\cos x)}{x^{2}}= 1 (2)

    ... and from (2)...

    \displaystyle \lim_{x \rightarrow 0} (1+\cos x)\ \lim_{x \rightarrow 0} \frac{1-\cos x}{x^{2}} = 1 \implies \frac{1-\cos x}{x^{2}}= \frac{1}{2} (3)

    Kind regards

    \chi \sigma
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