Calculating $\lim_{x\rightarrow 0}\frac{1}{x^n}e^{\frac{-1}{x^2}}$ (by hand)

Question

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

I tried using de l'Hospital but can't get any further

Thanks!

Answer 1

Let $\frac1{x}=y$ with $y\to \pm\infty$

$$\frac1{x^n}e^{\frac{-1}{x^2}}=\frac{y^n}{e^{y^2}}\to 0$$

indeed $\,\forall n$ eventually $e^{y^2}>y^{2n}$ and

$$0\le \frac{|y|^n}{e^{y^2}}\le \frac{|y|^n}{y^{2n}}\to 0$$

thus for squeeze theorem

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

Answer 2

$e^{u}>u^{n}$ for large $u>0$, so for small $|x|>0$, then $e^{-1/x^{2}}\leq x^{2n}$, so $\left|\dfrac{1}{x^{n}}e^{-1/x^{2}}\right|\leq\dfrac{|x|^{2n}}{|x|^{n}}=|x|^{n}\rightarrow 0$.

Answer 3

The key inequality is:

$$e^{y}\geq \frac{y^k}{k!}$$ for any $y\geq0$ and integer $k\geq 0.$ (This inequality follows directly fromn $e^y=\sum_{n=0}^{\infty}\frac{y^k}{k!}$, since all the terms in this sum are non-negative when $y\geq 0.$)

Letting $y=\frac{1}{x^2}$ you can deduce that:

$$0<\frac{e^{-1/x^2}}{x^{2k}}\leq k!$$

Then pick any $k$ so that $2k>n.$ So you get:

$$0<\frac{e^{-1/x^2}}{|x|^n}\leq |x|^{2k-n}k!$$

Then apply the squeeze theorem.

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This method proves more generally:

Claim: If $\{a_k\}$ is a sequence of non-negative numbers such that $f(x)=\sum_{k=0}^{\infty} a_kx^k$ converges for all $x\in \mathbb R$ then $$\lim_{x\to 0} \frac{1}{f(1/x^2)x^n}=0$$ if and only for some $k$, $2k>n$ and $a_k>0$.

Proof:

The "only if" part is easy - if there is no such $k$ then $f$ is a polynomial with $2\deg f\leq n.$ Then $g(x)=f(1/x^2)x^n$ is also a polynomial in $x$, and thus $x^nf(1/x^2)$ converges to a real value as $x\to 0.$

The "if" part is that $f(1/x^2)|x|^n\geq a_k|x|^{n-2k}>0$, so:

$$0<\frac{1}{f(1/x^2)|x|^n}\leq \frac{|x|^{n-2k}}{a_k}$$

and then the squeeze theorem is applied.

When $f$ is not a polynomial (so infinitely many $a_k>0)$ this means that there is always such a $k$ for any $n$ and thus the limit is zero for all $n.$