 June Bryan

2023-03-09

How to evaluate the limit $lim\frac{{e}^{t}}{t}$ as $t\to \infty$? pomorijujfo

Method 1: L'Hôpital's Rule
The limit:

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

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:

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

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

$\underset{t\to \infty }{lim}\frac{{e}^{t}}{t}=\underset{t\to \infty }{lim}\frac{\frac{d}{dt}{e}^{t}}{\frac{d}{dt}t}$
$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=\underset{t\to \infty }{lim}\frac{{e}^{t}}{1}$
= oo 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 $y=\frac{{e}^{x}}{x}$ is unbounded

Method 3: Taylor Series

So then:

$\underset{t\to \infty }{lim}\frac{{e}^{t}}{t}=\underset{t\to \infty }{lim}\left\{\frac{1}{t}+1+\frac{t}{2!}+\frac{{t}^{2}}{3!}+\frac{{t}^{3}}{4!}+...\right\}$
$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=\underset{t\to \infty }{lim}\left\{\frac{1}{t}+1+O\left(t\right)\right\}$
$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=\underset{t\to \infty }{lim}\left\{\frac{1}{t}+1\right\}+\underset{t\to \infty }{lim}\left\{O\left(t\right)\right\}$

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