# L'Hopital's Rule Help

• March 21st 2011, 12:35 PM
azncocoluver
L'Hopital's Rule Help
http://i71.photobucket.com/albums/i1...4/Untitled.jpg

Can someone please explain to me the red circle? I don't understand how the previous step results in 1/infinity divided by infinity which then equals to zero..?

Also, I don't understand anything in the picture below, so can someone also explain each step to me?

http://i71.photobucket.com/albums/i1...ionz04/4-1.jpg

Thank you!
• March 21st 2011, 12:50 PM
Also sprach Zarathustra
{2ln(x)*(1/x)}/{2x} = {ln(x)}/{x}*(1/x) --->(x-->infty) 0*0=0

In second problem:

n*sin(n)=sin(1/n)/{1/n} --->(n-->infty) sin(0)/(0)=0/0 {sinx is continuous function}

We ca use L'Hopital's rule! (WHY?)

{sin(1/n)}'/{1/n}'={cos(1/n)*(-1/n^2)}/{(-1/n^2}=cos(1/n) --->(n-->infty) cos(0)=1 {cosx is continuous function}
• March 21st 2011, 01:35 PM
azncocoluver
Thank you! I understand now, I just have one more question. In the first picture, 1/x is circled and infinity is plugged in for x. Why isn't ln x circled and infinity plugged in, as well?
• March 21st 2011, 03:55 PM
$\displaystyle\frac{lnx\left(\frac{1}{x}\right)}{x} =\frac{lnx}{x^2}$