find limit

**szpengchao** · February 28th 2008, 08:02 AM

limit ( e^(-1/x^2)/x,x,tends to infinity)

**colby2152** · February 28th 2008, 08:11 AM

Originally Posted by szpengchao

limit ( e^(-1/x^2)/x,x,tends to infinity)

It goes to zero. Why?

$x^2$ goes to infinity

$\frac{1}{x^2}$ then goes to zero

$e^{\frac{-1}{x^2}}$ approaches one

$\frac{1}{x}$ approaches zero

**ThePerfectHacker** · February 28th 2008, 08:37 AM

Maybe he meant $\lim_{x\to 0 }\frac{\exp (-1/x^2)}{x}$ .

**Krizalid** · February 28th 2008, 08:45 AM

Originally Posted by szpengchao

limit ( e^(-1/x^2)/x,x,tends to infinity)

By setting $u=\frac1{x^2}$ your limit is 0.

**galactus** · February 28th 2008, 10:44 AM

Here's a method. I hope it will suffice. Kind of one sided though.

$\lim_{x\rightarrow{0}}\frac{e^{\frac{-1}{x^{2}}}}{x}$

Let $t=e^{\frac{-1}{x^{2}}}$

Then you get:

$\lim_{t\rightarrow{0^{+}}}t\sqrt{-ln(t)}=0$

You can always try the "ol' Hospital" rule.

**xifentoozlerix** · February 28th 2008, 10:54 AM

L'Hôpital's rule seems to be a dead end for this problem. The second form ends up as $\mathop {\lim }\limits_{x \to 0} \frac{{( - 1)( - 2)x^{ - 3} e^{\frac{{ - 1}}{{x^2 }}} }}{1} \implies \mathop {\lim }\limits_{x \to 0} \frac{{2e^{\frac{{ - 1}}{{x^2 }}} }}{{x^3 }}$ . I don't know if there is any analytical way to solve this without the use of a substitution.

Thread: find limit

LinkBack

Thread Tools

Display

find this limit

Similar Math Help Forum Discussions

Find ∫ 1/(1-4x^2) dx with a low limit of 0 and a high limit of 1

How to find this limit?

Find the limit

1. find the limit 2. use the definition of the limit to prove that

How do I find this limit?

Search Tags