MHB 242.7x.27 Find the slowest growing and the fastest growing functions

[karush]

$\tiny{242.7x.27}$
$\textsf{Find the slowest growing and the fastest growing functions
${{x}\to{\infty}}$}$
\begin{align*}\displaystyle
y&=4x^{10} \\
y&=e^x \\
y&=e^{x-4} \\
y&=xe^x \\
\end{align*}

$\textit{I'm clueless... take the limit??}$

 

[MarkFL]

Yes, limits are involved here...

[box=green]Definition: Let $f(x)$ and $g(x)$ be positive for $x$ sufficiently large.

1. $f(x)$ grows faster than $g(x)$ as $x\to\infty$ if $$\lim_{x\to\infty}\frac{f(x)}{g(x)}=\infty$$
2. $f(x)$ grows slower than $g(x)$ as $x\to\infty$ if $$\lim_{x\to\infty}\frac{f(x)}{g(x)}=0$$
3. $f(x)$ and $g(x)$ grow at the same rate as $x\to\infty$ if $$\lim_{x\to\infty}\frac{f(x)}{g(x)}=L\ne0$$
   
   where $L$ is some finite number.​

[/box]

In order to compute the limits involved we often use L'Hôpital’s Rule. :D

 

[karush]

well that was very helpfull
but what determines what is f(x) and g(x)
what will these functions be compared to?

 

[MarkFL]

well that was very helpfull
but what determines what is f(x) and g(x)
what will these functions be compared to?

You determine which are to be $f$ and $g$...for example, if we compare the 2nd and 3rd options we could let:

$$f(x)=e^x$$

$$g(x)=e^{x-4}$$

And we find:

$$\frac{f(x)}{g(x)}=\frac{e^x}{e^{x-4}}=e^4$$

Hence:

$$\lim_{x\to\infty}\frac{f(x)}{g(x)}=e^4$$

So, we conclude that $f$ and $g$ grow at the same rate as $x\to\infty$. We would come to the same conclusion if we reversed the definitions of $f$ and $g$, as we would for functions who don't grow at the same rate as well. :D

 

[karush]

$\displaystyle
\lim_{{x}\to{\infty}}\frac{4x^{10}}{e^x}=0$
so this means $e^{x}$ is faster i presume..
however this is a calculator answer i wouldn't know how to evaluate it

also
$\displaystyle
\lim_{{x}\to{\infty}}\frac{e^{x-4}}{xe^x}=\textit{undef}$

so then ?

 

[Greg]

$\displaystyle
\lim_{{x}\to{\infty}}\frac{4x^{10}}{e^x}=0$
so this means $e^{x}$ is faster i presume..
however this is a calculator answer i wouldn't know how to evaluate it

Try repeated applications of L'Hopital's rule, until the $x$ in the numerator vanishes.

also
$\displaystyle
\lim_{{x}\to{\infty}}\frac{e^{x-4}}{xe^x}=\textit{undef}$

so then ?

This limit is actually $0$.

 

[HallsofIvy]

$\displaystyle
\lim_{{x}\to{\infty}}\frac{4x^{10}}{e^x}=0$
so this means $e^{x}$ is faster i presume..
however this is a calculator answer i wouldn't know how to evaluate it

also
$\displaystyle \lim_{{x}\to{\infty}}\frac{e^{x-4}}{xe^x}=\textit{undef}$

so then ?

[math]\frac{e^{x-4}}{xe^x}= \frac{e^xe^{-4}}{xe^x}= \frac{e^{-4}}{x}[/math]

That is a constant over x so, as x goes to infinity, the numerator stays constant while the denominator increases without bound, so the limit is 0.