June Bryan

2023-03-09

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

pomorijujfo

Beginner2023-03-10Added 7 answers

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

$\frac{{e}^{t}}{t}=\frac{1}{t}{e}^{t}$

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

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

So then:

$\underset{t\to \infty}{lim}\frac{{e}^{t}}{t}=\underset{t\to \infty}{lim}\{\frac{1}{t}+1+\frac{t}{2!}+\frac{{t}^{2}}{3!}+\frac{{t}^{3}}{4!}+...\}$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=\underset{t\to \infty}{lim}\{\frac{1}{t}+1+O\left(t\right)\}$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=\underset{t\to \infty}{lim}\{\frac{1}{t}+1\}+\underset{t\to \infty}{lim}\left\{O\left(t\right)\right\}$

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

$\frac{{e}^{t}}{t}=\frac{1}{t}{e}^{t}$

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

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

So then:

$\underset{t\to \infty}{lim}\frac{{e}^{t}}{t}=\underset{t\to \infty}{lim}\{\frac{1}{t}+1+\frac{t}{2!}+\frac{{t}^{2}}{3!}+\frac{{t}^{3}}{4!}+...\}$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=\underset{t\to \infty}{lim}\{\frac{1}{t}+1+O\left(t\right)\}$

$\phantom{\rule{1ex}{0ex}}\text{}\phantom{\rule{1ex}{0ex}}=\underset{t\to \infty}{lim}\{\frac{1}{t}+1\}+\underset{t\to \infty}{lim}\left\{O\left(t\right)\right\}$

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