Puzzled by this application of L'Hospital's rule

[quasar987]

[SOLVED] Puzzled by this application of L'Hospital's rule

Homework Statement

In an example in my textbook, they define a function f on the real line by f(0)=0 and f(x)=exp(-1/x^2) otherwise.

They then say that we can evaluate f'(0) by L'Hospital's rule, and they write

f'(0)=\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x-0}=\lim_{x\rightarrow 0}\frac{e^{-1/x^2}}{x}=0

How did they get that?? It seems if I apply L'Hospital, I get

\lim_{x\rightarrow 0}\frac{e^{-1/x^2}}{x}=\lim_{x\rightarrow 0}\frac{-2x^{-3}e^{-1/x^2}}{1}=\lim_{x\rightarrow 0}\frac{-2e^{-1/x^2}}{x^{3}}

and thus I'm not more advanced!

 

[rock.freak667]

Well I guess you can say that as x tends to zero

\frac{1}{x^2}\rightarrow \infty faster than \frac{1}{x}
so that exp(-1/x^2) tends to zero before 1/x (from the denominator) can tend to infinity.

 

[eok20]

\lim_{x\rightarrow 0}\frac{e^{-1/x^2}}{x}= \lim_{x\rightarrow 0} \frac{x^{-1}}{e^{1/x^2}} and now do L'Hospital's rule.

 

[Dick]

Change the variable to s=1/x. Now you have the limit as s->infinity of s/e^(s^2). Now use l'Hopital. Which is what eok is basically saying.