Evaluating \lim_{\ x \to 0^+} \frac{\ e^{-1/x}}{x}

[qspeechc]

Homework Statement

Evaluate: $$\lim_{\ x \to 0^+} \frac{\ e^{-1/x}}{x}$$

Homework Equations

l'hospital's' Rule maybe? Taylor series?

The Attempt at a Solution

Before I did anything, my guess was this tends to 0, as exponentials decay faster than rational funtions.

The limits as -1/x tends to 0 from the right is $$-\infty$$
So:
$$\lim_{\ x \to 0^+} \ e^{-1/x} = 0$$

and ofcourse the limit of x as x tends to 0 is 0. So it is of indeterminate form 0/0
Using L'Hopital's Rule:

$$\lim_{\ x \to 0^+} \frac{\ e^{-1/x}}{x^2}$$

And that is still of indeterminate form 0/0. In fact, you can see that using l'hospital's Rule will always generate an indeterminate form 0/0. Taylor series are no better: after the exapansion you get
$$\frac{1}{x^2} +\frac{1}{2x^5} +...$$ or something of that nature, point is, it diverges.
Help please?

 

[D H]

Look at it from the other end of the reals: Let $u=1/x$. Then

$$\frac{e^{-\;1/x}}{x} = ue^{-u} = \frac{u}{e^u}$$

 

[arildno]

You might take the logarithm of the fraction, and try to evaluate:
$$\lim_{x\to0^{+}}-\frac{1}{x}-\ln(x)=\lim_{x\to0^{+}}-\frac{1+x\ln(x)}{x}$$

 

[Sourabh N]

try subs x = 1/ ln y and then try L'hopital..

 

[Sourabh N]

thats correct.

 

[qspeechc]

Thank-you everyone for your answers, but I do not like substitutions, like put x=(...),
is there any way to solve it without the substitution?

Look at it from the other end of the reals:

I do not get this substitution, or any other substitution for that matter! Why, and how, look at it from the other end? Or is it merely a substitution one must use to solve the problem, but has no real meaning. Am I making sense?

 

[D H]

Making substitutions is a very important tool. You had better learn to love them if you want to proceed far in mathematics.

In the case of my example, u=1/x, the substitution changes an indeterminate expression of the form $0/0$ as $x\to0$ to an indeterminate expression f the form $\infty/\infty$ as $u\to\infty$. The advantage of the substitution is that using L'Hopital's Rule yields a value.

 

[qspeechc]

Ok, thank you again everyone, otherwise I would have really struggled with this one!