Failure of L’Hôpital’s Rule: $ \lim\limits_{x\rightarrow 0^+} \dfrac{e^{-1/x}}{x^p} $ Consider
$$
\lim_{x\rightarrow 0^+} \frac{e^{-1/x}}{x^p}
$$
where $p\in \mathbb{R}^+\;(p>0)\;.$
Despite satisfying all of the criteria for L’Hôpital’s Rule, repeated applications continue to yield the same indeterminate limit, whereas it can be shown heuristically, that the limit is indeed zero.
I don't see how the Squeeze Theorem would help me here. Is there an alternative? Thanks.
Many thanks for all the answers. I wish I'd thought of the clever $u=1/x$ substitution beforehand!
 A: You can still use it, after a change of variable $u=1/x$ that
\begin{align*}
\lim_{u\rightarrow\infty}\dfrac{u^{p}}{e^{u}}.
\end{align*}
Since $p>0$, repeating the L'Hopital $m$ times such that $p-m<0$ will do.
A: $\frac{d}{dx} e^{-1/x} = e^{-1/x}\cdot \frac{1}{x^2}$
$\frac{d}{dx} x^p = p x^{p-1}$
So L'Hopital's Rule gives
$$\lim_{x\rightarrow 0^+} \frac{e^{-\frac{1}{x}}}{x^p} = \lim_{x\rightarrow 0^+} \frac{e^{-\frac{1}{x}}}{px^{p+1}}$$
and working backwards we could get down to a single $x$ in the denominator:$$\lim_{x\rightarrow 0^+} \frac{e^{-\frac{1}{x}}}{x} = \lim_{x\rightarrow 0^+} \frac{e^{-\frac{1}{x}}}{(p-1)!x^{p}}$$
If we let $u = \frac{1}{x}$, the limit converts to
$$\lim_{x\rightarrow 0^+} \frac{e^{-\frac{1}{x}}}{x} = \lim_{u\rightarrow +\infty} u{e^{-u}} = \lim_{u\rightarrow +\infty} \frac{u}{e^u} = \lim_{u\rightarrow +\infty} \frac{1}{e^u} = 0$$
A: Let substitute $u=\frac 1x\to+\infty$ then $$f(u)=u^pe^{-u}$$
$f'(u)=-u^{p-1}e^{-u}(u-p)<0$ because $u>0$, exp is always positive and $u-p>0$ for $u$ large enough.
We have $f(u)\ge 0$ and $\searrow$ at infinity so it is bounded therefore
$$f(u)=2^p\left(\tfrac u2\right)^pe^{-\frac u2}e^{-\frac u2}=\underbrace{2^p}_\text{cst}\underbrace{f(\tfrac u2)}_\text{bounded}\underbrace{e^{-\frac u2}}_{\to 0}\to 0$$
A: Calculate $\;\lim\limits_{x\rightarrow 0^+}\dfrac{e^{-1/x}}{x^p}\;$
where $\;p\in \mathbb{R}^+\;(p>0)\;.$
We are going to calculate the limit by using the Squeeze Theorem.
Since $\;u<e^u\;$ for all $\;u\in\mathbb{R}\;,\;$ it results that
$0<\dfrac1{2px}<e^{1/(2px)}\;$ for all $\;x\in\mathbb{R}^+\;,$
$0<\dfrac1{(2p)^px^p}<e^{1/(2x)}\;$ for all $\;x\in\mathbb{R}^+\;,$
$0<\dfrac{e^{-1/x}}{(2p)^px^p}<e^{-1/(2x)}\;$ for all $\;x\in\mathbb{R}^+\;,$
$0<\dfrac{e^{-1/x}}{x^p}<(2p)^pe^{-1/(2x)}\;$ for all $\;x\in\mathbb{R}^+\;.$
Given that $\;\lim\limits_{x\to0^+}(2p)^pe^{-1/(2x)}=0\;,\;$ by using the Squeeze Theorem, it follows that
$\lim\limits_{x\rightarrow 0^+}\dfrac{e^{-1/x}}{x^p}=0\;.$
A: You can understand the function $e^{-1/x}$ as having a root at $x=0$ of infinite multiplicity, that's what makes L'Hospital fail. At the same time this is a heuristic argument to justify the answer $0$, as the denominator $x^p$ cannot compensate.
Taking the logarithm,
$$-x-p\log x\sim-x$$ is also conclusive.
