# Puzzled by this application of L'Hospital's rule

1. Jun 5, 2008

### quasar987

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

1. The problem statement, all variables and given/known data
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!

2. Jun 5, 2008

### 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.

3. Jun 5, 2008

### 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.

Last edited: Jun 5, 2008
4. Jun 5, 2008

### 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.