## chrisplusian 2 years ago find the limit of (n^p)/(e^n) as n approaches infinity

1. ParthKohli

What is $$p$$?

2. chrisplusian

P>0 I really dont even understand why they give that information accept that it shows n remains in the numerator

3. chrisplusian

I tried to use logarithmic properties to find an answer but it just kept running me in circles

4. ParthKohli

$n^p \times \frac{1}{e^{n}}$$$e$$ is a constant. Oh, and $$n^p \to \infty$$ and $$\dfrac{1}{e^n} \to 0$$. That can only mean that their product is approaching 0.

5. ParthKohli

Am I right? O_O

6. terenzreignz

Definitely not the reasoning I'd pick...

7. ParthKohli

Knew it.

8. experimentX

correct answer but wrong logic ... think about the opposite e^n/n^p and n->infinity

9. terenzreignz

I prefer simple cases... $\Large 2e^n \rightarrow \infty$$\Large \frac1{e^n}\rightarrow 0$ $\Large 2e^n \times \frac1{e^n }\rightarrow 0 \quad \color{red}?$

10. chrisplusian

See i don't understand becasue if you look at it as (N^p)*(1/e^n) then the n^p is infinity and the !/e^n is zero. That gives you infinity times zero which is an indeterminate form.

11. DLS

0

12. chrisplusian

Sorry (1/e^n)

13. terenzreignz

Relax... you can either use L'Hôpital or one of my favourite theorems~ pick one

14. ParthKohli

Approaching infinity, not really infinity. I tried to use l'hopital: didn't work. :-|

15. terenzreignz

The almighty Squeeze~

16. DLS

17. ParthKohli

ARMAGERD. How could I forget the squeeze

18. chrisplusian

Ok I tried lopitals and it did not work nor did logarithmic properties

19. ParthKohli

What's the inequality BTW?

20. chrisplusian

21. DLS

Yes,its easy.

22. DLS

Use LH

23. ParthKohli

@DLS how?

24. experimentX

My favouraite technique is $a = e^{\log a} \\ n^{p} = e^{p \log n }$

25. terenzreignz

I realise that L'Hôpital may not work after all... at least, not yet... Use this... Let $$\large q = \lceil p\rceil$$ such that... $\Large 0 \le\frac{n^p}{e^n}\le \frac{n^q}{e^n}$

26. terenzreignz

q is guaranteed to be a positive integer, means it [the numerator] will (eventually) differentiate to zero...

27. chrisplusian

@experimentX why would that help?

28. experimentX

L'hopital works use it least floor(p + 1)

29. terenzreignz

Why deviate from the norm, @experimentX ? :D Just use ceiling :3

30. DLS

Well observe the denominator. Its e^n.No matter how many times you differentiate it,i.e use L-Hopsital,the denominator WON'T CHANGE at all.That is the first thing.Now I say,use L hospital infinity times(not really) so if you keep on differentiating..a time will come when numerator turns out to be 0. so 0/1=1. Example: $y=5x^4$ $\frac{dy}{dx}=20x^3=>60x^2=>120x=>120$

31. chrisplusian

Don't understand what you mean @experimentX

32. experimentX

@chrisplusian |dw:1369325101292:dw|

33. terenzreignz

To simplify things, we best make sure the exponent p in the numerator is an (positive, but that's granted) integer. Taking $$\Large q = \lceil p\rceil$$ helps with that.

34. terenzreignz

And then squeezing :)

35. experimentX

@terenzreignz yes ceil

36. DLS

@chrisplusian my reply should probably clear it off. @experimentX does it seem right?

37. chrisplusian

the solutions guide says =0 when p>0, and bless than or equal to 2