Lim x->0 e^(-1/x^2)/x^3 (Don't Understand Why I Can't Do It!)

  • #1

Homework Statement



It is part of a larger problem, but the only hangup I have had is computing this limit.
lim x->0 e^(-1/x^2)/x^3.

It's to show that the function f(x) = e^(-1/x^2) (when x is not 0) and 0 (when x is 0) is not equal to its Maclurin Series. I know that if I can show that the derivatives of f(x) are equal to zero when x=0 (a) then I can prove what the problem is asking. Showing why the derivatives equal zero has been a problem thus far though...

Homework Equations





The Attempt at a Solution



I put it into Wolfram Alpha and know that it's equal to zero, but I don't know how to get it.
I know that L'Hopital's rule applies because it is 0/0
so:
((2/x^3)*e^(-1/x^2))/(3x^2) = 2e^(-1/x^2)/3x^5
Doesn't this pattern keep repeating giving one 0/0 no matter how many times one differentiated? It will always have e^(-1/x^2) on the top and some power of x which will continue to grow on the bottom. How can this limit be computed? Thanks!
 

Answers and Replies

  • #2
rock.freak667
Homework Helper
6,230
31
As [itex]x \rightarrow 0[/itex] which goes to zero first? e-1/x2 or x3?
 
  • #3
I don't quite understand what you're asking; Both go to zero and I don't understand how one can do so first...
 
  • #4
392
0
[tex]\frac{e^{-1/x^2}}{x^3}=\frac{x^{-3}}{e^{1/x^2}}[/tex]

Try l'Hopital twice on that instead, simplifying after each application.
 
  • #5
Thanks! Is it possible to generalize that e^(-1/x^2)/x^n as n->0 (n>2 and an integer) also approaches 0?
 

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