1. Question about applying L'Hospitals rule

Hi,

Say I have a limit that, on a first direct evaluation, gives (0 x infinity)/0. Can I still use L'Hospitals rule? The specific limit I've been given is (x^2(sin(1/x)))/sinx as x approaches 0. It seems that I could divide the numerator by 1/sin(1/x)), which would give 0/0/0 which doesn't seem to help. Is it even possible?

Thanks!

3. Re: Question about applying L'Hospitals rule

Hi,

$\displaystyle x^2\sin(1/x)$ does approach 0 as x approaches 0. One way to see this:

$\displaystyle |\sin(z)|\leq1$ for any z, so $\displaystyle 0\leq|x^2\sin(1/x)\leq|x^2|$ for x not 0. The squeeze theorem says the limit is 0.

So you can try l'hopital's rule by differentiating both numerator and denominator. However, when you do this you get a limit which does not exist. So to find the limit, you must try something else. You should check on a possible limit by graphing the function and zooming in towards the origin.

Post again if you still can't figure it out.