Question

Find horizontal asymptote of \(f(x) =5x^3e^{x}\)
Please, help

Answer 1

@SolomonZelman

Answer 2

I would like to know
1) the name of the property:

\(lim_{x\rightarrow-\infty} f(x) =\lim_{x\rightarrow \infty}f(-x)\)

2) Is there any way else to find the horizontal asymptote without using that property??

Answer 3

By definition, \(\color{black}{f(x)=\beta }\) is horizontal asymptote iff \(\color{black}{x\to \pm \infty}\) at \(\color{black}{f(x)=\beta}\).

Answer 4

So, going from the definition, your horizontal asymptote is \(\color{black}{f(x)=0}\), because \(\color{black}{x\to -\infty }\) at that point. (Although, this is sure not like the "typical" asymptote case.)

Answer 5

Logic, please.

Answer 6

How can you get 0?

Answer 7

\(\color{black}{\lim_{x\to-\infty}5x^3e^x=0}\)

Answer 8

We have \(lim_{x\rightarrow \infty}\dfrac{x^n}{e^x}=0\)

So, we have to apply the property 1) above to get that form. Right?

Answer 9

Hey, how do you ask questions?
Really need help with a question!

Answer 10

\(\color{black}{\displaystyle \lim_{x\to-\infty}5x^3e^x=\lim_{x\to\infty}\frac{-5x^3}{e^x}=-5\lim_{x\to\infty}\frac{x^3}{e^x}}=0\)

if you want it in that form, ... but even without this, we can surely tell the limit is 0.

Answer 11

Common, I need know the name of that property. Not just "trust me!!, we can surely get it"

Answer 12

That the line \(\color{black}{\displaystyle y=b}\) is a horizontal asymptote of \(\color{black}{\displaystyle f}\) if \(\color{black}{\displaystyle \lim_{x\to \pm \infty}f(x)=b}\).

Answer 13

You need the name of this property? I don't have it ...

Answer 14

Sometimes, you have y = b is a horizontal asymptote of f from both sides \(\pm \infty\)

Sometimes, we have just one way. not both

Answer 15

Yes, that is true ...

Answer 16

And there is no way to find the limit when x approaches to -infinitive, we have to turn it to +infinitive and change the f(x) to f(-x)

Unfortunately, not all of profs give the students that property.

Answer 17

oh you want the "official" way of solving that limit?

Answer 18

Yes, please

Answer 19

\(\color{black}{\displaystyle \lim_{x\to -\infty} 5x^3e^x}\)
\(\tiny \\[0.9em]\)
Substitution: \(\color{black}{\displaystyle x=-p}\)\(\tiny \\[0.2em]\)
Then, \(\color{black}{\displaystyle 5x^3=-5p^3{\tiny~}}\) and \(\color{black}{\displaystyle e^x=e^{-p}=\frac{1}{e^p}{\tiny~}}\) (by direct substitution).\(\tiny \\[0.9em]\)
Also, as \(\color{black}{\displaystyle x\to -\infty }\), then \(\color{black}{\displaystyle -p\to -\infty }\), and therefore \(\color{black}{\displaystyle p\to +\infty }\).

Thus, your new limit is
\(\color{black}{\displaystyle \lim_{p\to \infty} \frac{-5p^3}{e^p}=-5\lim_{p\to \infty} \frac{p^3}{e^p}}\)

and knowing that \(\color{black}{\displaystyle \forall n,m\in\mathbb{R}\left[m\times \lim_{p\to \infty} \frac{p^n}{e^p}=0\right]}\), we get \(\color{black}{\displaystyle -5\lim_{p\to \infty} \frac{p^3}{e^p}=0}\).

Answer 20

((The most important thing in this "official" is the substitution, really, and that we apply the property of the limit at infinity, of a polynomial over exponentially increasing function ... any more rigorousness is just a spice))

Answer 21

Thank you so much. I got it.

Answer 22

Just didn't know what exactly you were asking to back up:) YW

Answer 23

\(\color{black}{\displaystyle \lim_{z\to \infty} \frac{e^x}{\Gamma(z)}=0}\) :)

Answer 24

jk

Answer 25

oh no, gamma function!!

Answer 26

You don't like the gamma function?