- #1

joypav

- 151

- 0

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)