anonymous
  • anonymous
plz help! evaluate this:
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
|dw:1443422490893:dw|
Jhannybean
  • Jhannybean
\[\large \lim_{x\rightarrow \infty} x^3e^{-x/5}\] think you can just plug in \(\infty\) after making both the functions positive to see the trend
Jhannybean
  • Jhannybean
\[\large \lim_{x\rightarrow \infty} \dfrac{x^3}{e^{x/5}}\] so plugging in \(\infty\), we see that \[\lim_{x\rightarrow \infty} \frac{\infty^3}{e^{\infty}/5} = 0\] \(e^{-x/5}\) will just reach infinity a lot faster than x\(^3\)

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Astrophysics
  • Astrophysics
You can apply L'hopital's rule
Jhannybean
  • Jhannybean
But how would LH help with the denominator?
Astrophysics
  • Astrophysics
Because you have \[\lim_{x \rightarrow \infty} \frac{ x^3 }{ e^{\frac{ x }{ 5 }} }\]
Astrophysics
  • Astrophysics
Oh you can apply it more than once
Astrophysics
  • Astrophysics
Probably would have to a few times xD
Astrophysics
  • Astrophysics
I'm just waiting for oldrins amazing explanation
anonymous
  • anonymous
0?
Astrophysics
  • Astrophysics
Right!
anonymous
  • anonymous
thanks guys!!!!:-)
Astrophysics
  • Astrophysics
@magy33 wait though, oldrin and ganeshie might have a better explanation/ easier way of doing it
anonymous
  • anonymous
consider $$e^{x/5}=1+\frac{x}5+\frac{x^2}{50}+\frac{x^3}{750}+O(x^4)\\\implies e^{-x/5}=\frac1{1+x/5+x^2/50+x^3/750+o(x^4)}\\\implies x^3e^{-x/5}=\frac{x^3}{1+x/5+x^2/50+x^3/750+o(x^4)}\\\qquad\qquad\qquad=\frac1{1/x^3+1/5x^2+1/50x+1/750+o(x)}$$ as \(x\to\infty\) now in the limit \(x\to\infty\) we have that \(1/x^3,1/5x^2,1/50x\to0\) giving $$x^3e^{-x/5}\sim\frac1{1/750+o(x)}\to0$$ in other words we say that \(x^3e^{-x/5}=x^3/e^{x/5}\to 0\) or equivalently \(x^3\in o(e^{x/5})\), i.e. \(e^{x/5}\) *dominates* \(x^3\) in the limit as \(x\to\infty\). like Jhanny said, \(e^{x/5}\) a heck of a lot faster than \(x^3\) overall, and quickly outpaces it forcing the limit to zero
ganeshie8
  • ganeshie8
I still have this unanswered question : In one of the lectures, professor Mattuck says that the limit \(\lim\limits_{x\to\infty}\dfrac{x^n}{e^x}\) can be evaluated by applying LH \(n\) times. He also says that if you're clever enough, it can also be evaluated by applying LH just one time. But he wont tell how to do that...
anonymous
  • anonymous
a more formal way of showing this is in terms of l'Hopital's rule, which is used to compare the growth rates of functions in infinite limits
ganeshie8
  • ganeshie8
Probably, he was refering to taylor too...
Jhannybean
  • Jhannybean
Im not sure how to apply L'Hospital's rule to this... And how did you simplify x\(^3\)/(expansion) \(\rightarrow\) x/(1/expansion) ? I think that's what happened there...
Jhannybean
  • Jhannybean
sorry, typo, 1/(1/expansion) *
anonymous
  • anonymous
$$\frac{x^n}{e^x}=n^n\left(\frac{x/n}{e^{x/n}}\right)^n$$ so consider passing the limit in: $$\lim_{x\to\infty} n^n\left(\frac{x/n}{e^{x/n}}\right)^n=n^n\left(\lim_{u\to\infty}\frac{u}{e^u}\right)^n$$ with \(u=x/n\)
anonymous
  • anonymous
@ ganeshie8
ganeshie8
  • ganeshie8
Wow, brilliant! pretty sure that is it! I had been waiting so long... but that idea has never struck me, really clever trick!
ganeshie8
  • ganeshie8
https://youtu.be/sZ2qulI6GEk?t=2599
Astrophysics
  • Astrophysics
Lol wow, I was excited, so now I know who professor Mattuck is, leaving with a cliffhanger. Oldrin's trick looks clever, but I don't entirely understand it xD
ganeshie8
  • ganeshie8
$$\lim\limits_{x\to\infty}\frac{x^n}{e^x}=\lim\limits_{x\to\infty}n^n\left(\frac{x/n}{e^{x/n}}\right)^n$$ with \(u=x/n\) : $$\lim_{x\to\infty} n^n\left(\frac{x/n}{e^{x/n}}\right)^n=n^n\left(\lim_{u\to\infty}\frac{u}{e^u}\right)^n=n^n(0)^n = 0$$
ganeshie8
  • ganeshie8
Notice that we had to apply LH only one time, not "n" times...
ganeshie8
  • ganeshie8
that is the clever part :)
Astrophysics
  • Astrophysics
The taylor series method he also showed was pretty cool

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