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Can you please show me how to continue? I don't know how to continue.

$$\lim_{x \rightarrow 0} e^{-\frac{1}{x^2}}\cdot \frac{2}{x^3}$$

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Change variable: $t=1/x^2$, so that $t \to +\infty$. – Siminore Mar 12 '14 at 12:27
@Siminore $t=1/x$ would be better though, I don't like radicals. – Sabyasachi Mar 12 '14 at 12:28
But then you should, in principle, split $x \to 0+$ and $x \to 0-$... – Siminore Mar 12 '14 at 12:29
so you are suggesting that I compute $\lim_{x \rightarrow \infty} e^{-t}*2t\sqrt{t}$? – user134934 Mar 12 '14 at 12:31

2 Answers 2

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For this type of problem, a helpful substitution is $t = \frac{1}{x}$. Observe that when $x \to 0$, $t \to \infty$. Thus, your limit becomes $$ \lim_{t \to \infty} e^{-t^2} 2t^3.$$ You may proceed from here by L'Hospital's rule.

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thank you, but what about the @Siminore remark, that we have to consider $x \rightarrow 0_+$ and $x \rightarrow 0_-$? Or it doesn't matter? – user134934 Mar 12 '14 at 12:39
If you try both cases ($t \to +\infty$ and $t \to -\infty$), you will see they both come out to the same answer. – David Zhang Mar 12 '14 at 12:42
I see, I tried it and you are indeed correct, thank you once again. – user134934 Mar 12 '14 at 12:44

We wish to compute the limit,

$$\lim_{x\to 0}\, \, 2\frac{e^{-x^{-2}}}{x^3}.$$

Perform a substitution $t=x^{-2}$, such that the limit becomes as $t\to +\infty$, i.e.

$$\lim_{t\to +\infty} 2 t^{3/2}e^{-t} =0 .$$

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