Attachment 25372

I know this goes to zero as u goes to zero from my calculator, but how does one break this down algebraically?

Printable View

- Oct 23rd 2012, 03:00 PMkingsolomonsgravehow to evaluate this algebraically
Attachment 25372

I know this goes to zero as u goes to zero from my calculator, but how does one break this down algebraically? - Oct 23rd 2012, 03:12 PMrichard1234Re: how to evaluate this algebraically
L'Hôpital's rule:

$\displaystyle \lim_{u \to 0} \frac{\sin u^2}{u} = \lim_{u \to 0} \frac{2u \cos u^2}{1} = 0$ - Oct 24th 2012, 07:21 AMSironRe: how to evaluate this algebraically
Without l'Hopital's rule you can use the fact that $\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x} = 1$

We have

$\displaystyle \lim_{u \to 0} \frac{\sin u^2}{u} = \lim_{u \to 0} \frac{\sin u^2}{u} \frac{u}{u} = \lim_{u \to 0} \frac{\sin u^2}{u^2} \lim_{u \to 0} u$

If $\displaystyle u \to 0$ then $\displaystyle u^2 \to 0$ thus $\displaystyle \lim_{u \to 0} \frac{\sin u^2}{u^2} = 1$ and $\displaystyle \lim_{u \to 0} u = 0$ therefore

$\displaystyle \lim_{u \to 0} \frac{\sin u^2}{u} = 0$