Limit of $\dfrac{e^{-1/|x|}}{x^n}$ as $x\to 0$

Question

I need to prove that for all $n\in\mathbb{N}$ the next equality holds: $$\lim\limits_{x\to 0}\dfrac{e^{-1/|x|}}{x^n}=0$$First, I thought in induction. If $n=0$ then $\lim\limits_{x\to 0}\dfrac{e^{-1/|x|}}{x^n}=\lim\limits_{x\to 0}{e^{-1/|x|}}=0$. If we suppose that $\lim\limits_{x\to 0}\dfrac{e^{-1/|x|}}{x^n}=0$ then we want to prove that $\lim\limits_{x\to 0}\dfrac{e^{-1/|x|}}{x^{n+1}}=0$ but this is equal to $\lim\limits_{x\to 0}\dfrac{e^{-1/|x|}}{x^n}\cdot\dfrac{1}{x}$ but here the limit of $1/x$ when $x$ approaches to zero doesn't exists. Then I can't use the product of the limits. I think that L'Hôpital doesn't works because we have an absolute value in the function. Any hint? I really appreciate any help you can provide me.

Answer 1

Let assume wlog $x>0$ and $y=1/x\to \infty$ then by l'Hopital

• Base case n=1

$$\lim\limits_{x\to 0}\dfrac{e^{-1/x}}{x}=\lim\limits_{y\to \infty}\dfrac{y}{e^{y}}\stackrel{H.R.}=\lim\limits_{y\to \infty} \dfrac{1}{e^{y}}=0$$

• Induction step

assume

$$\lim\limits_{x\to 0}\dfrac{e^{-1/x}}{x^{n}}=0$$

then

$$\lim\limits_{x\to 0}\dfrac{e^{-1/x}}{x^{n+1}}=\lim\limits_{y\to \infty}\dfrac{y^{n+1}}{e^{y}}\stackrel{H.R.}=\lim\limits_{y\to \infty}(n+1)\dfrac{y^{n}}{e^{y}}=\lim\limits_{x\to 0}(n+1)\dfrac{e^{-1/x}}{x^{n}}=0$$

Answer 2

Make the change: $y=\frac 1x$. Evaluate the left hand and right hand limits: $$\lim\limits_{x\to 0+}\dfrac{e^{-1/|x|}}{x^n}=\lim\limits_{x\to 0+}\dfrac{e^{-1/x}}{x^n}=\lim\limits_{y\to +\infty}\dfrac{e^{-y}}{\frac{1}{y^n}}=\lim\limits_{y\to +\infty}\dfrac{y^n}{e^{y}}=0,$$ because the exponential function grows faster than a polynomial function of any degree. $$\lim\limits_{x\to 0-}\dfrac{e^{-1/|x|}}{x^n}=\lim\limits_{x\to 0-}\dfrac{e^{1/x}}{x^n}=\lim\limits_{y\to -\infty}\dfrac{e^{y}}{\frac{1}{y^n}}=\lim\limits_{y\to -\infty}\dfrac{y^n}{e^{-y}}=0,$$ because again the exponential function grows faster than a polynomial function of any degree.