How do I calculate this limit

Question

$$\lim_{x\to 0 } \,\,\frac{e^{-1/x^2}}{x^6}$$

I tried doing it with substitution and L'hospital but it doesn't help. I also tried expanding it with Taylor series and calculating the limit and the result is not the same as the one I get using Wollfam Alpha.

Answer 1

You can show the limit using squeezing and the series expansion for $e^u$:

• $(\star)$: $e^u = \sum_{n=0}^{\infty}\frac{u^n}{n!} \stackrel{}{>}\frac{u^4}{4!} $ for $u > 0$

Using this with $u = \frac 1{x^2}$ you get

$$0 \leq \frac{e^{-1/x^2}}{x^6}= \frac{1}{x^6e^{\frac{1}{x^2}}}\stackrel{(\star)}{\leq} \frac{1}{x^6\frac{1}{4!x^8}}= 4!x^2\stackrel{x\to 0}{\longrightarrow} 0$$

Answer 2

The given limit is same as $\lim_{x \to \infty} \frac {x^{6}} {e^{x^{2}}}$ or or $\lim_{y \to \infty} \frac {y^{3}} {e^{y}}$. Use L'Hopital's Rule to show that the limit is $0$.