MHB Evaluate Limit: $\frac{0}{0}$ - Can Someone Help?

[joypav]

I'm trying to show...

$\lim_{{x}\to{0^+}}\left(\frac{e^(\frac{-1}{x})}{x^n}\right)=0$

I guess my calculus is a bit rusty. Can someone help me out?
Here's what I've got.

$\lim_{{x}\to{0^+}}ln\left(\frac{e^(\frac{-1}{x})}{x^n}\right)=\lim_{{x}\to{0^+}}\left(-\frac{1}{x}-ln(x^n)\right)=-\infty+\infty$

So, I put it in a form to get $\frac{0}{0}$ so I can use L'Hospital's.

$\lim_{{x}\to{0^+}}\frac{x+\frac{1}{ln(x^n)}}{-\frac{x}{ln(x^n)}}$

Is this how I am supposed to be proceeding? (Wondering)
[/size]

 

[Greg]

There's a trick to this one.

-(1/x + nlog(x))

-(1/x - nlog(1/x))

-(u - nlog(u)) as u approaches infinity (a swap with 1/x) :)

-(1 - nlog(u)/u)/(1/u)

Hence we have (if you'll pardon the notation ) $e^{-\infty}=0$.[/size]

I'm assuming $n$ is a constant.

 

[joypav]

$-(u - nlog(u))$

$\frac{-(1-nlog(u)}{\frac{1}{u}}$

How do you get from here to here?
I understand that if you factor out a $u$ the first term becomes $1$, but why does $nlog(u)$ stay the same?

 

[Greg]

My apologies - i made an error. I have edited my post.

 

[joypav]

My apologies - i made an error. I have edited my post.

Gotcha. Thanks!