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anonymous

  • one year ago

plz help! evaluate this:

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  1. anonymous
    • one year ago
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    |dw:1443422490893:dw|

  2. Jhannybean
    • one year ago
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    \[\large \lim_{x\rightarrow \infty} x^3e^{-x/5}\] think you can just plug in \(\infty\) after making both the functions positive to see the trend

  3. Jhannybean
    • one year ago
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    \[\large \lim_{x\rightarrow \infty} \dfrac{x^3}{e^{x/5}}\] 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\(^3\)

  4. Astrophysics
    • one year ago
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    You can apply L'hopital's rule

  5. Jhannybean
    • one year ago
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    But how would LH help with the denominator?

  6. Astrophysics
    • one year ago
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    Because you have \[\lim_{x \rightarrow \infty} \frac{ x^3 }{ e^{\frac{ x }{ 5 }} }\]

  7. Astrophysics
    • one year ago
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    Oh you can apply it more than once

  8. Astrophysics
    • one year ago
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    Probably would have to a few times xD

  9. Astrophysics
    • one year ago
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    I'm just waiting for oldrins amazing explanation

  10. anonymous
    • one year ago
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    0?

  11. Astrophysics
    • one year ago
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    Right!

  12. anonymous
    • one year ago
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    thanks guys!!!!:-)

  13. Astrophysics
    • one year ago
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    @magy33 wait though, oldrin and ganeshie might have a better explanation/ easier way of doing it

  14. anonymous
    • one year ago
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    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

  15. ganeshie8
    • one year ago
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    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...

  16. anonymous
    • one year ago
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    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

  17. ganeshie8
    • one year ago
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    Probably, he was refering to taylor too...

  18. Jhannybean
    • one year ago
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    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...

  19. Jhannybean
    • one year ago
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    sorry, typo, 1/(1/expansion) *

  20. anonymous
    • one year ago
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    $$\frac{x^n}{e^x}=n^n\left(\frac{x/n}{e^{x/n}}\right)^n$$ 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 \(u=x/n\)

  21. anonymous
    • one year ago
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    @ ganeshie8

  22. ganeshie8
    • one year ago
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    Wow, brilliant! pretty sure that is it! I had been waiting so long... but that idea has never struck me, really clever trick!

  23. ganeshie8
    • one year ago
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    https://youtu.be/sZ2qulI6GEk?t=2599

  24. Astrophysics
    • one year ago
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    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

  25. ganeshie8
    • one year ago
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    $$\lim\limits_{x\to\infty}\frac{x^n}{e^x}=\lim\limits_{x\to\infty}n^n\left(\frac{x/n}{e^{x/n}}\right)^n$$ with \(u=x/n\) : $$\lim_{x\to\infty} n^n\left(\frac{x/n}{e^{x/n}}\right)^n=n^n\left(\lim_{u\to\infty}\frac{u}{e^u}\right)^n=n^n(0)^n = 0$$

  26. ganeshie8
    • one year ago
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    Notice that we had to apply LH only one time, not "n" times...

  27. ganeshie8
    • one year ago
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    that is the clever part :)

  28. Astrophysics
    • one year ago
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    The taylor series method he also showed was pretty cool

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is replying to Can someone tell me what button the professor is hitting...

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