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

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1. anonymous

|dw:1443422490893:dw|

2. anonymous

$\large \lim_{x\rightarrow \infty} x^3e^{-x/5}$ think you can just plug in $$\infty$$ after making both the functions positive to see the trend

3. anonymous

$\large \lim_{x\rightarrow \infty} \dfrac{x^3}{e^{x/5}}$ 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

You can apply L'hopital's rule

5. anonymous

But how would LH help with the denominator?

6. Astrophysics

Because you have $\lim_{x \rightarrow \infty} \frac{ x^3 }{ e^{\frac{ x }{ 5 }} }$

7. Astrophysics

Oh you can apply it more than once

8. Astrophysics

Probably would have to a few times xD

9. Astrophysics

I'm just waiting for oldrins amazing explanation

10. anonymous

0?

11. Astrophysics

Right!

12. anonymous

thanks guys!!!!:-)

13. Astrophysics

@magy33 wait though, oldrin and ganeshie might have a better explanation/ easier way of doing it

14. 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

15. 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...

16. 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

17. ganeshie8

Probably, he was refering to taylor too...

18. anonymous

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. anonymous

sorry, typo, 1/(1/expansion) *

20. 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 $$u=x/n$$

21. anonymous

@ ganeshie8

22. ganeshie8

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
24. 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

25. 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 $$u=x/n$$ : $$\lim_{x\to\infty} n^n\left(\frac{x/n}{e^{x/n}}\right)^n=n^n\left(\lim_{u\to\infty}\frac{u}{e^u}\right)^n=n^n(0)^n = 0$$

26. ganeshie8

Notice that we had to apply LH only one time, not "n" times...

27. ganeshie8

that is the clever part :)

28. Astrophysics

The taylor series method he also showed was pretty cool