# Limit with parameter, $(e^{-1/x^2})/x^a$

How do I explore for which values of the parameter $a$

$$\lim_{x\to 0} \frac{1}{x^a} e^{-1/x^2} = 0?$$ For $a=0$ it is true, but I don't know what other values.

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## 3 Answers

Hint: You may rewrite the equation

$$\frac{1}{x^a} e^{\frac{-1}{x^2}} = \frac{1}{x^a e^{\frac{1}{x^2}}} = \frac{1}{e^{a \ln(x)+ \frac{1}{x^2}}}$$

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but now we need to calculate $\lim _{x \to 0}a\ln x +\frac{1}{x^2}$. I don't think it's a simpler limit to figure out. –  Amihai Zivan Jan 11 '13 at 12:16

It's true for all $a's$ (note that for $a\leqslant 0$ it's trivial) and you can prove it easily using L'Hopital's rule or Taylor's series for $\exp(-x^2)$.

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Hint: Put $y=1/x$ and note that $y \to \infty$.

$$\lim_{y\to \infty} y^a e^{-y^2} = \dots.$$

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