Proof: limit=0 for any positive integer n

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dincerekin
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Homework Statement


Prove that [tex]\lim_{x\to0}\frac{e^\frac{-1}{x^2}}{x^n}=0[/tex] for any positive integer n.

Homework Equations





The Attempt at a Solution


I've tried using a combination of induction and l'hopital's rule to no avail. Perhaps I am over complicating it?

All help is appreciated

Thanks in advance
 
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I applied l'hospital's and simplified a bit and now I've got

[tex]\frac{n}{2}\lim_{x\to0}\frac{x^2}{e^\frac{1}{x^2}x^n}=0[/tex]

now what?
 
so I am trying to show its true for n=k+1 assuming n=k

i.e i need to show that [tex]\lim_{x\to0}\frac{x^{1-k}}{e^\frac{1}{x^2}}=0[/tex]

assuming that [tex]\lim_{x\to0}\frac{x^{2-k}}{e^\frac{1}{x^2}}=0[/tex]

im not sure how to manipulate this now?
 
dincerekin said:
so I am trying to show its true for n=k+1 assuming n=k

i.e i need to show that [tex]\lim_{x\to0}\frac{x^{1-k}}{e^\frac{1}{x^2}}=0[/tex]

assuming that [tex]\lim_{x\to0}\frac{x^{2-k}}{e^\frac{1}{x^2}}=0[/tex]

im not sure how to manipulate this now?

Split the first term into the one you know(assumed), and its factor. Then apply basic limits.
 
so,

[tex]\lim_{x\to0}\frac{x^{2-k}}{e^\frac{1}{x^2}}=\lim_{x\to0}\frac{x^{-1}x^{1-k}}{e^\frac{1}{x^2}}[/tex]

but i can't simply say that
[tex]\lim_{x\to0}\frac{x^{-1}x^{1-k}}{e^\frac{1}{x^2}}= \lim_{x\to0}{x^{-1}} × \lim_{x\to0}\frac{x^{1-k}}{e^\frac{1}{x^2}}[/tex]

because [tex]\lim_{x\to0}{x^{-1}}[/tex] doesn't exist, right?
 
dincerekin said:
so,

[tex]\lim_{x\to0}\frac{x^{2-k}}{e^\frac{1}{x^2}}=\lim_{x\to0}\frac{x^{-1}x^{1-k}}{e^\frac{1}{x^2}}[/tex]

Uhh, this is wrong. Does multiplying x-1 and x1-k give x2-k??
 
oh sorry, that should be the other way around
 
it should be
[tex]\lim_{x\to0}\frac{x^{1-k}}{e^\frac{1}{x^2}}=\lim_{x\to0}\frac{x^{-1}x^{2-k}}{e^\frac{1}{x^2}}[/tex]
 
micromass said:
Uuuh, or slightly more useful

[tex]\lim_{x\rightarrow 0}\frac{x^{2-k}}{e^{1/x^2}}=\lim_{x\rightarrow 0}\frac{x\cdot x^{1-k}}{e^{1/x^2}}[/tex]

Can you split up the limits now?
oh! how didn't I see this before

thanks so much <3