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Puzzled by this application of L'Hospital's rule

  • Thread starter quasar987
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  • #1
quasar987
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[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

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

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

[tex]\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}}[/tex]

and thus I'm not more advanced!
 

Answers and Replies

  • #2
rock.freak667
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Well I guess you can say that as x tends to zero

[itex]\frac{1}{x^2}\rightarrow \infty[/itex] faster than [itex]\frac{1}{x}[/itex]
so that exp(-1/x^2) tends to zero before 1/x (from the denominator) can tend to infinity.
 
  • #3
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[itex] \lim_{x\rightarrow 0}\frac{e^{-1/x^2}}{x}= \lim_{x\rightarrow 0} \frac{x^{-1}}{e^{1/x^2}} [/itex] and now do L'Hospital's rule.
 
Last edited:
  • #4
Dick
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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.
 

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