plz help! evaluate this:

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plz help! evaluate this:

Calculus1
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|dw:1443422490893:dw|
\[\large \lim_{x\rightarrow \infty} x^3e^{-x/5}\] think you can just plug in \(\infty\) after making both the functions positive to see the trend
\[\large \lim_{x\rightarrow \infty} \dfrac{x^3}{e^{x/5}}\] 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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You can apply L'hopital's rule
But how would LH help with the denominator?
Because you have \[\lim_{x \rightarrow \infty} \frac{ x^3 }{ e^{\frac{ x }{ 5 }} }\]
Oh you can apply it more than once
Probably would have to a few times xD
I'm just waiting for oldrins amazing explanation
0?
Right!
thanks guys!!!!:-)
@magy33 wait though, oldrin and ganeshie might have a better explanation/ easier way of doing it
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
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...
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
Probably, he was refering to taylor too...
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...
sorry, typo, 1/(1/expansion) *
$$\frac{x^n}{e^x}=n^n\left(\frac{x/n}{e^{x/n}}\right)^n$$ 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
Wow, brilliant! pretty sure that is it! I had been waiting so long... but that idea has never struck me, really clever trick!
https://youtu.be/sZ2qulI6GEk?t=2599
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
$$\lim\limits_{x\to\infty}\frac{x^n}{e^x}=\lim\limits_{x\to\infty}n^n\left(\frac{x/n}{e^{x/n}}\right)^n$$ with \(u=x/n\) : $$\lim_{x\to\infty} n^n\left(\frac{x/n}{e^{x/n}}\right)^n=n^n\left(\lim_{u\to\infty}\frac{u}{e^u}\right)^n=n^n(0)^n = 0$$
Notice that we had to apply LH only one time, not "n" times...
that is the clever part :)
The taylor series method he also showed was pretty cool

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