# How do you evaluate the limit lim e^t/t as t->oo?

Feb 25, 2017

Method 1: L'Hôpital's Rule

The limit:

${\lim}_{t \rightarrow \infty} {e}^{t} / t$

is of an indeterminate form $\frac{\infty}{\infty}$, and so we can apply L'Hôpital's rule which states that for an indeterminate limit then, providing the limits exits then:

${\lim}_{x \rightarrow a} f \frac{x}{g} \left(x\right) = {\lim}_{x \rightarrow a} \frac{f ' \left(x\right)}{g ' \left(x\right)}$

And so applying L'Hôpital's rule we get:

${\lim}_{t \rightarrow \infty} {e}^{t} / t = {\lim}_{t \rightarrow \infty} \frac{\frac{d}{\mathrm{dt}} {e}^{t}}{\frac{d}{\mathrm{dt}} t}$

$\text{ } = {\lim}_{t \rightarrow \infty} \frac{{e}^{t}}{1}$
$\text{ } = \infty \setminus \setminus \setminus \setminus$ as ${e}^{t}$ is unbounded.

Method 2: Graphically

graph{(e^x)/x [-26.97, 23.66, -2.97, 22.35]}

Nothing more to say; clearly $y = {e}^{x} / x$ is unbounded

Method 3: Taylor Series

${e}^{t} / t = \frac{1}{t} \setminus {e}^{t}$
 \ \ \ \ = 1/t {1+t+t^2/(2!) + t^3/(3!) + t^4/(4!) + ... }
 \ \ \ \ = 1/t+1+t/(2!) + t^2/(3!) + t^3/(4!) + ...

So then:

 lim_(t rarr oo) e^t/t = lim_(t rarr oo) {1/t+1+t/(2!) + t^2/(3!) + t^3/(4!) + ... }
$\text{ } = {\lim}_{t \rightarrow \infty} \left\{\frac{1}{t} + 1 + O \left(t\right)\right\}$
$\text{ } = {\lim}_{t \rightarrow \infty} \left\{\frac{1}{t} + 1\right\} + {\lim}_{t \rightarrow \infty} \left\{O \left(t\right)\right\}$

Although the first limit is finite, the second is unbounded.