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

How would I go about calculating this limit?

Thankyou

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- Nov 1st 2010, 03:36 AMkatedew987Calculating a Limit
lim x --> 0 ((1-cos x)/ x^2)

How would I go about calculating this limit?

Thankyou - Nov 1st 2010, 03:48 AMUnknown008
Maybe you can use the l'Hopital's rule twice.

That is, you differentiate the numerator and denominator with respect to x twice. - Nov 1st 2010, 03:51 AMkatedew987
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!

- Nov 1st 2010, 05:03 AMharish21
- Nov 1st 2010, 05:06 AMkatedew987
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

- Nov 1st 2010, 01:53 PMhjortur
How did you do the original proof that

$\displaystyle \lim\limits_{x\to0} \frac{1-cos(x)}{x}=0$ ? - Nov 1st 2010, 02:02 PMAlso sprach Zarathustra
Use $\displaystyle 2sin^2(\frac{x}{2})=1-cos(x)$

And use:

YOU will get that the limit is 1/2 - Nov 1st 2010, 02:08 PMDrexel28
- Nov 1st 2010, 02:16 PMchisigma
Starting from the 'fundamental limit'...

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

... You first derive...

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

... and from (2)...

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

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$