1. Calculating a Limit

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

How would I go about calculating this limit?

Thankyou

2. Maybe you can use the l'Hopital's rule twice.

That is, you differentiate the numerator and denominator with respect to x twice.

3. 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!

4. 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 $\neq 0$

5. 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

6. How did you do the original proof that
$\lim\limits_{x\to0} \frac{1-cos(x)}{x}=0$ ?

7. Use $2sin^2(\frac{x}{2})=1-cos(x)$

And use:

YOU will get that the limit is 1/2

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

How would I go about calculating this limit?

Thankyou
$\displaystyle \lim_{x\to 0}\frac{1-\cos(x)}{x^2}=\lim_{x\to 0}\left(\frac{\sin(x)}{x}\right)^2\cdot\frac{1}{1+ \cos(x)}$

9. 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

$\chi$ $\sigma$