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How can I calculate the following limit,

$$\lim_{x\to0}\frac{e^{-1/x^2}}x=0$$

If I plug $x=0$ I get the following,

$$f'(0)=\lim_{x\to0}\frac{e^{-1/0^2}}0=\infty$$

jh123
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  • That the value of its derivative doesn't exist doesn't imply the limit doesn't exist. By the way, are you sure what you calculate is its derivative $f'(x)?$ It just looks like that you plug in the value $0.$ – User Mar 17 '17 at 16:04

1 Answers1

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Let $x=1/u$. Then we have

$$\lim_{x\to0}\frac{e^{-1/x^2}}{x}=\lim_{u\to\pm\infty}\frac{e^{-u^2}}{1/u}$$

Now apply L'Hospital's rule to get

$$\lim_{u\to\pm\infty}\frac{e^{-u^2}}{1/u}=\lim_{u\to\pm\infty}e^{-u^2}=e^{-\infty}=0$$

Simply Beautiful Art
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