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anonymous
 one year ago
plz help! evaluate this:
anonymous
 one year ago
plz help! evaluate this:

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443422490893:dw

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1\[\large \lim_{x\rightarrow \infty} x^3e^{x/5}\] think you can just plug in \(\infty\) after making both the functions positive to see the trend

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1\[\large \lim_{x\rightarrow \infty} \dfrac{x^3}{e^{x/5}}\] 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\(^3\)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2You can apply L'hopital's rule

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1But how would LH help with the denominator?

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Because you have \[\lim_{x \rightarrow \infty} \frac{ x^3 }{ e^{\frac{ x }{ 5 }} }\]

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Oh you can apply it more than once

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Probably would have to a few times xD

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2I'm just waiting for oldrins amazing explanation

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2@magy33 wait though, oldrin and ganeshie might have a better explanation/ easier way of doing it

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0consider $$e^{x/5}=1+\frac{x}5+\frac{x^2}{50}+\frac{x^3}{750}+O(x^4)\\\implies e^{x/5}=\frac1{1+x/5+x^2/50+x^3/750+o(x^4)}\\\implies x^3e^{x/5}=\frac{x^3}{1+x/5+x^2/50+x^3/750+o(x^4)}\\\qquad\qquad\qquad=\frac1{1/x^3+1/5x^2+1/50x+1/750+o(x)}$$ 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
 one year ago
Best ResponseYou've already chosen the best response.2I 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
 one year ago
Best ResponseYou've already chosen the best response.0a 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
 one year ago
Best ResponseYou've already chosen the best response.2Probably, he was refering to taylor too...

Jhannybean
 one year ago
Best ResponseYou've already chosen the best response.1Im 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
 one year ago
Best ResponseYou've already chosen the best response.1sorry, typo, 1/(1/expansion) *

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$\frac{x^n}{e^x}=n^n\left(\frac{x/n}{e^{x/n}}\right)^n$$ 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 \(u=x/n\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2Wow, brilliant! pretty sure that is it! I had been waiting so long... but that idea has never struck me, really clever trick!

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2Lol 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
 one year ago
Best ResponseYou've already chosen the best response.2$$\lim\limits_{x\to\infty}\frac{x^n}{e^x}=\lim\limits_{x\to\infty}n^n\left(\frac{x/n}{e^{x/n}}\right)^n$$ 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
 one year ago
Best ResponseYou've already chosen the best response.2Notice that we had to apply LH only one time, not "n" times...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.2that is the clever part :)

Astrophysics
 one year ago
Best ResponseYou've already chosen the best response.2The taylor series method he also showed was pretty cool
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