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#1 2006-09-23 01:00:57

Finney
Guest

Limits

Hello ,

while studing limits of sequences , I've found 2 main formulas to be very difficult to solve. The book does not show the proof but states that :

1 - Lim ( n ---> infinity ) X^n = 0 if     -1 < X < 1
I found out that the limit for X <= -1 should not exist , I couldn't determine the limit for -1 < X < 0 ( but the book says it's 0 without proof ) and I found out that the limit is infinity if 1 < X and the lim is 1 if X = 1. Can anyone confirm this and explain why the lim is 0 for -1 < x < 0 ?

2- Lim ( n ---> infinity ) X^n / n! = 0    for any X
Why ? How can I use L'hopital's rule here if I dont know what the derivative of n! is w.r.t n ?

Thanks

#2 2006-09-23 03:12:22

Malik641
Member
Registered: 2005-10-18
Posts: 3

Re: Limits

Hey,

I think that for #1 the concept is that the higher the power is, the smaller the number becomes. It eventually becomes so small, it can be considered to be 0. Whether X is -1 < X < 0 -or- 0 < X < 1.

The same thing goes for #2. Try putting some large values for n. You'll find the larger you go, the smaller the number becomes. Like 2^99 / 99! = 6.79150329946 x 10^-127 This is VERY small. So small, it can be considered zero.


Hope this helps

Last edited by Malik641 (2006-09-23 03:13:10)

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#3 2006-09-23 03:42:29

Dross
Member
Registered: 2006-08-24
Posts: 325

Re: Limits

First off, how detailed a description do you want for answers to these questions? Here are some descriptive/appeal to intuition-type examples. If you'd like a

proof, just ask.

First off, note that (if you're considering integer values of n) for any x < 0:

- the (-1)[sup]n[/sup] part just makes sure we "alternate" between positive and negative values.

Now you can see that

, because the sequence
for -1<x<0 is the same as the sequence for
when 0<x<1, except it "alternates" between positive and negative values - the important thing is that it still, regardless of sign, gets arbitrarily close to 0.

More formally, you could use the theorem:

(for a

proof, just write out the definition of each and you'll soon see they are equivalent)

Now, using the theorem we can write:


As for your second question,

because n! is "more powerful" than x[sup]n[/sup].

Think of the series

. As the series progresses, you multiply the numerator by x and the denominator by n each time. No matter how large the x that you choose is, the n in the series will - eventually - surpass it. You will then be multiplying the numerator by a smaller number than the denominator (so, multiplying your series by a number between 0 and 1) forever more, so the series will eventually settle down to 0 again.


Is this all clear? If not, feel free to ask again!

Last edited by Dross (2006-09-24 11:02:11)


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#4 2006-09-24 07:37:42

Finney
Guest

Re: Limits

Malik641 ,
In the first part what seems illogical to me is how a -ve number to the power infinity exists when its sign depends on the power itself ( even - odd ) . So how can the limit exist if x is negative ? It makes no sense to me. For the 2nd part , yes i know it does make sense but can u prove it without trial and error?

Dross ,
Well frankly , I cant say that I understood  the 1st part. I cant just cant see how a certain function alternates between 2 values when n ---> infinity and still has a limit = 0 , I thought in this case the limit should not exist , which is quite reasonable to me. I mean what's the limit of (-1)^n when n---> infinity ? Should it not exist ?

I've also encountered the delta - epsilon proof somewhere in the book , but I found it to be very illogical , so I dont really use it.

Concerning the 2nd part , yes ur right , it does make sense that eventually the limit would be 0 , but I thought if asked such a thing , I need to prove it algebraicly or something , right ?

Thank you two for your time.

#5 2006-09-24 08:58:20

luca-deltodesco
Member
Registered: 2006-05-05
Posts: 1,470

Re: Limits

Finney wrote:

I cant just cant see how a certain function alternates between 2 values when n ---> infinity and still has a limit = 0 , I thought in this case the limit should not exist , which is quite reasonable to me.

usually, you would be right, if a function alternates between 2 values, there is no limit to infinity, however, in this case, both the two values that the function alternates between, diverge to the same value, so in this case you can give a limit


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#6 2006-09-24 11:09:49

Dross
Member
Registered: 2006-08-24
Posts: 325

Re: Limits

Finney wrote:

Well frankly , I cant say that I understood  the 1st part. I cant just cant see how a certain function alternates between 2 values when n ---> infinity and still has a limit = 0 , I thought in this case the limit should not exist , which is quite reasonable to me. I mean what's the limit of (-1)^n when n---> infinity ? Should it not exist ?

Ah, but the function does not alternate between two values - it alternates between positive and negative. The important thing, however (as demonstrated by the short theorem I put up in my last post, namely:

)

...is that the size of the values decreaces. Thus, if you go far enough along the x-axis, the value of your function gets arbitrarily close to zero.

Also, you would be wise to pay an awful lot of attention to delta-epsilon proofs and definitions. For years, the concept of a limit was not very well defined, and this caused a lot of trouble. The delta-epsilon definition of a limit is very important - it is the bedrock of limits, and limits are the bedrock of calculus. No, that was an understatement - limits are calculus, and the delta-epsilon definition of a limit is what makes it mathematicaly rigorous, as opposed to just some vague statement, tossed about without care.


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#7 2006-09-28 09:49:35

Finney
Guest

Re: Limits

Hi again ,

Ah, but the function does not alternate between two values - it alternates between positive and negative.

I think I got what you mean. It's like the limit alternates between +0.00001 and -0.00001 and hence we can safely say the limit converges to 0 , correct ?

Now concerning the Epsion-delta proof , I never really payed attention it to primarily because I never found anything logical in it , nor has the book seriously used this method to prove anything serious. Anyway , I will re-check the theory and ask about it ASAP.

Thank you for your time.

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