Help me in finding the limit

1. Jun 21, 2007

macjack

limit n->infinity ((-1)^(n-1))/n ?

i tried to find the limit using the limit at infinity trick,...i got the value as 0
but the solution given are not matched.

Thanks
Mc

2. Jun 21, 2007

ice109

limit doesn't exist

3. Jun 21, 2007

Dick

The limit is zero. If the question is misstated and meant to be a series sum then its not.

4. Jun 22, 2007

ice109

limit oscillates? limits that oscillate don't exist?

5. Jun 22, 2007

VietDao29

What do you mean by "Limit oscillate"? The limit is 0, it does not oscillate. However, the value for the expression does oscillate, but, in the end, they all converge to 0 so the limit does exist, and it's 0. It's like the limit:
$$\lim_{n \rightarrow \infty} \frac{\sin n}{n} = 0$$
The value for the expression oscillate around 0, but as n tends to infinity, they converge to 0, so the limit is 0.

Not to be confused with, $$\lim_{n \rightarrow \infty} (-1) ^ n$$. This limit does not exist, since, when n is odd, the value for the expression is -1, and when it's even, it's 1. They don't converge, so there's no limit there.

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@macjack: Are you familiar with the Squeeze Theorem?

Last edited: Jun 22, 2007
6. Jun 22, 2007

macjack

Thanks

a) The sequence approaches the limit from RHS
b) The sequence approaches the limit from LHS
c) The sequence oscillates about the limit
d) none of the above.

None of my answers and few of them answers also didnt match this
choose.

and in the answer section he didnt mention anything about which one is the correct one...it just says...sequence:1, -1/2, 1/3......

what does it mean ?? is that question wrongly stated ?? i have this doubt because he is start using 1,2,3...for 'n' right ??
any clue ??

I don't know about the squeeze theorem !! Let me check out that today.

Thanks

7. Jun 22, 2007

ice109

dude... you have to tell people it's a sequence. in which case the correct answer is c

8. Jun 22, 2007

macjack

thanks ice109

thanks ...
i am a newbie into calculus..didnt face the limits of sequences till now.

9. Jun 22, 2007

ice109

now someone help me, why is the limit of the expression 0 and not nonexistant

10. Jun 22, 2007

Dick

Because the difference between zero and (-1)^(n-1)/n approaches zero as n approaches infinity. The sense of the multiple choices is simply how does it approach it. But approach it, it certainly does. 1/n goes to zero and the sign oscillation doesn't change that.

11. Jun 22, 2007

macjack

why i said it is 0 is ?

there is a common trick when you take lim x->infinity, based on the powers of the numerator and denominator in a polynomial equation.

1) if the orders are same , then just take the coefficients of higher order terms and divide it.

2) if the order of denominator is bigger than the top, then its limit is 0.

3) if the order of the denominator is lesser than the top, then the limit is infinity
or no limit (i have a doubt here)...am i correct with this condition..

so for this problem, we satified second condition..so i said it is 0...

12. Jun 22, 2007

Dick

It is zero. Are you doubting it?

13. Jun 23, 2007

VietDao29

Yes, you can do it this way. When n tends to infinity, the denominator also grows without bound, so +1, or -1, when divided by that denominator will tend to 0. So, the limit is 0.

14. Jun 23, 2007

macjack

what about the third statement ?

>3) if the order of the denominator is lesser than the top, then the limit is >infinity or no limit (i have a doubt here)...am i correct with this condition..

Did anyone know what is the correct answer for this ??

15. Jun 23, 2007

CompuChip

As Dick already said, the limit can be defined as follows: a sequence $$(x_n)_{n \in \mathbb{N}}$$ has a limit x, if we can make the difference between xn and x as small as we want by choosing n appropriate.

For example, we see that $$x_n = 1/n$$ has limit zero, because if I give you some number $$\epsilon > 0$$ you can always give me an n for which $$1/n < \epsilon$$. On the other hand, $$x_n = \frac{n}{2}$$ does not have a limit. For example, if you claim it's "infinity", I can ask you for a number n such that $$|\frac{n}{2} - \infty| < \epsilon$$ and you will be unable to give it to me, however big my epsilon is, since the distance is always infinity. In this case, we say the limit does not exist, although to indicate this we usually (technically speaking, quite sloppily) write
$$\lim_{n \to \infty} \frac{n}{2} = \infty$$.

Now maybe it's a nice exercise for you to try and prove that way the limit (or its non-existence) of
$$\frac{x^n}{x^m}$$
for the cases n < m, n = m and n > m.

16. Jun 23, 2007

HallsofIvy

Staff Emeritus
IF the sequence {an} converges to a non-zero number, then the sequence {(-1)nan} does not converge since, if we assume a limit of a, some terms of the sequence, for arbitrarily large n, will be at least |a| (distance from a to 0) away. If, however, {an} converges to 0, {(-1)nan} also converges to 0.