1. ## find this limit

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

2. Originally Posted by szpengchao
limit ( e^(-1/x^2)/x,x,tends to infinity)
It goes to zero. Why?

$\displaystyle x^2$ goes to infinity

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

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

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

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

4. Originally Posted by szpengchao
limit ( e^(-1/x^2)/x,x,tends to infinity)
By setting $\displaystyle u=\frac1{x^2}$ your limit is 0.

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

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

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

Then you get:

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

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

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