I want to find the limit $$\lim_{x\to\infty} \frac{e^{-x}}{x^{-1}}$$
Both sides go to 0, so I want to use L’Hospital’s Rule. However, when I differentiate both sides, I get $$\lim_{x\to\infty} \frac{-e^{-x}}{-x^{-2}}$$
Further differentiation doesn't yield anything useful either. How could I go about finding this limit?
On
Without L'Hopital: We know that $e^{x}\geq x^{2}$ for large $x>0$, so $\dfrac{x}{e^{x}}\leq\dfrac{1}{x}$, but $1/x\rightarrow 0$ as $x\rightarrow\infty$, so the limit is zero by Squeeze Theorem.
On
By definition, $e^x := \displaystyle \sum_{n=0}^\infty \frac{x^n}{n!}$, assuming that it converges.
For $x>0$, we see that the series consists of strictly positive real numbers, so $e^x > \displaystyle \sum_{n=0}^2 \frac{x^n}{n!} = 1+x+\frac{x^2}2 > \frac{x^2}2$.
Therefore, $\dfrac x {e^x} < \dfrac x {x^2/2} = \dfrac 2 x$.
To establish that $\displaystyle \lim_{x \to \infty} \frac x {e^x} = 0$, we need to see that $\dfrac x {e^x}$ becomes arbitrarily close to $0$ as $x$ increases without bound. By that, I mean that if you want $\dfrac x {e^x}$ to be at most $\varepsilon$ away from $0$, where $\varepsilon$ is any positive real number, it suffices to let $x > \dfrac 1 {2\varepsilon}$.
Indeed, if $x > \dfrac 1 {2\varepsilon}$, then $0 < \dfrac x {e^x} < \dfrac x {x^2/2} = \dfrac 2 x < \varepsilon$.
Notice
$$ \lim \frac{ e^{-x} }{x^{-1} } = \lim \frac{ x }{e^x} = \lim \frac{1}{e^x} = 0 $$