How to evaluate the limit lim e t t as t → ∞ ?

Method 1: L'Hôpital's Rule
The limit:

${\displaystyle \underset{t\to \infty }{lim}\frac{e^t}{t}}$

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

${\displaystyle \underset{x\to a}{lim}\frac{f\left(x\right)}{g\left(x\right)}=\underset{x\to a}{lim}\frac{f\prime \left(x\right)}{g\prime \left(x\right)}}$

Applying L'Hôpital's rule, we thus arrive at:

${\displaystyle \underset{t\to \infty }{lim}\frac{e^t}{t}=\underset{t\to \infty }{lim}\frac{\frac{d}{dt}{e}^t}{\frac{d}{dt}t}}$
${\displaystyle =\underset{t\to \infty }{lim}\frac{e^t}{1}}$
= oo as ${\displaystyle e^t}$ is unbounded.


Method 2: Graphically
graph{(e^x)/x [-26.97, 23.66, -2.97, 22.35]}
Nothing more to say; clearly ${\displaystyle y=\frac{e^x}{x}}$ is unbounded



Method 3: Taylor Series

${\displaystyle \frac{e^t}{t}=\frac{1}{t} e^t}$
${\displaystyle =\frac{1}{t}\{1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+\frac{t^4}{4!}+...\}}$
${\displaystyle =\frac{1}{t}+1+\frac{t}{2!}+\frac{t^2}{3!}+\frac{t^3}{4!}+...}$

So then:

${\displaystyle \underset{t\to \infty }{lim}\frac{e^t}{t}=\underset{t\to \infty }{lim}\{\frac{1}{t}+1+\frac{t}{2!}+\frac{t^2}{3!}+\frac{t^3}{4!}+...\}}$
${\displaystyle =\underset{t\to \infty }{lim}\{\frac{1}{t}+1+O\left(t\right)\}}$
${\displaystyle =\underset{t\to \infty }{lim}\{\frac{1}{t}+1\}+\underset{t\to \infty }{lim}\left\{O\left(t\right)\right\}}$