# Calculating a Limit

• Nov 1st 2010, 04:36 AM
katedew987
Calculating a Limit
lim x --> 0 ((1-cos x)/ x^2)

How would I go about calculating this limit?

Thankyou
• Nov 1st 2010, 04:48 AM
Unknown008
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, 04:51 AM
katedew987
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, 06:03 AM
harish21
Quote:

Originally Posted by katedew987
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 $\neq 0$
• Nov 1st 2010, 06:06 AM
katedew987
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, 02:53 PM
hjortur
How did you do the original proof that
$\lim\limits_{x\to0} \frac{1-cos(x)}{x}=0$ ?
• Nov 1st 2010, 03:02 PM
Also sprach Zarathustra
Use $2sin^2(\frac{x}{2})=1-cos(x)$

And use:

YOU will get that the limit is 1/2
• Nov 1st 2010, 03:08 PM
Drexel28
Quote:

Originally Posted by katedew987
lim x --> 0 ((1-cos x)/ x^2)

How would I go about calculating this limit?

Thankyou

$\displaystyle \lim_{x\to 0}\frac{1-\cos(x)}{x^2}=\lim_{x\to 0}\left(\frac{\sin(x)}{x}\right)^2\cdot\frac{1}{1+ \cos(x)}$
• Nov 1st 2010, 03:16 PM
chisigma
Quote:

Originally Posted by katedew987
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

$\chi$ $\sigma$