L'Hopital's Rule

[lamaclass]

I was just wondering if someone could tell me if I'm on the right track with this problem and how to finish it? Thanks!

Use L'Hopital's Rule to evaluate the limit.

lim xe[sup:3qfuo35e]-x^{[/sup:3qfuo35e]2} = -oo/oo
x--->oo

=ln y=ln x-x[sup:3qfuo35e]2[/sup:3qfuo35e]

=ln y=-x[sup:3qfuo35e]2[/sup:3qfuo35e]/1/ln x

After this, not sure how to continue from here.

 

[galactus]

Please fix your post. What is that?.

Is it supposed to be \(\displaystyle \lim_{x\to \infty}xe^{-x^{2}}\)

If so, rewrite \(\displaystyle \lim_{x\to \infty}\frac{x}{e^{x^{2}}}\), then apply the Hospital rule.

No one can answer if they can not read your post. :?:

 

[lamaclass]

Thank you very much Galactus! And yep it was supposed to be -x[sup:3119v5a4]2[/sup:3119v5a4] as a power but somehow it came out wrong. Thanks again!