Limit of $\lim_{x\to 0-} x^4e^{-1/x}$

Question

How can I calculate the limit of $\lim_{x\to 0-} x^4e^{-1/x}$

I tried using de l'Hospital but couldn't get and further because I always get undefined expressions. Thank you!

Answer 1

\begin{align*} \lim_{x\to0^-}x^4e^{\frac{-1}{x}}&=\lim_{x\to0^-}\frac{e^{\frac{-1}{x}}}{x^{-4}}\\ &=\lim_{x\to0^-}\frac{\frac{1}{x^2}e^{\frac{-1}{x}}}{-4x^{-5}}\\ &=\lim_{x\to0^-}\frac{e^{\frac{-1}{x}}}{-4x^{-3}}\\ &=\lim_{x\to0^-}\frac{\frac{1}{x^2}e^{\frac{-1}{x}}}{12x^{-4}}\\ &=\lim_{x\to0^-}\frac{e^{\frac{-1}{x}}}{12x^{-2}}\\ &=\lim_{x\to0^-}\frac{\frac{1}{x^2}e^{\frac{-1}{x}}}{-24x^{-3}}\\ &=\lim_{x\to0^-}\frac{e^{\frac{-1}{x}}}{-24x^{-1}}\\ &=\lim_{x\to0^-}\frac{\frac{1}{x^2}e^{\frac{-1}{x}}}{24x^{-2}}\\ &=\lim_{x\to0^-}\frac{e^{\frac{-1}{x}}}{24}\\ &=\infty \end{align*}

Answer 2

Set $-1/x=h$

to find $$\lim_{x\to0^-}x^4e^{-1/x}=\lim_{h\to\infty^+}\dfrac{e^h}{h^4}$$

Now apply L'Hôpital's Rule

Answer 3

Let $y:=-1/x$ and consider $y \rightarrow +\infty$, i.e.

$\lim_{y \rightarrow + \infty} \dfrac{e^y}{y^4}.$

$e^y \gt $

$1+y+y^2/2! +.....+ y^5/5! $

The limit is?

Answer 4

Note that by ratio test

$$a_n=\frac{e^n}{n^4}\implies\frac{a_{n+1}}{a_n}=e\cdot\left(1+\frac1n\right)^4\to e\implies a_n \to +\infty$$