Gill Math: Finding the Limit of (-1)^(n-1)/n using Infinity Trick

[macjack]

can you please help me to find the limit ?

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.

can you please help me out ?

Thanks
Mc

 

[ice109]

limit doesn't exist

 

[Dick]

limit doesn't exist

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

 

[ice109]

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

limit oscillates? limits that oscillate don't exist?

 

[VietDao29]

limit oscillates? limits that oscillate don't exist?

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?

 

[macjack]

Thanks

Thanks for your response.
The answers given are
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

 

[ice109]

Thanks for your response.
The answers given are
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

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

 

[macjack]

thanks ice109

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

 

[ice109]

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

 

[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.

 

[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...

 

[Dick]

It is zero. Are you doubting it?

 

[VietDao29]

...
2) if the order of denominator is bigger than the top, then its limit is 0.
...
so for this problem, we satified second condition..so i said it is 0...

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.

 

[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 ??

 

[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 x_n 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.

 

[HallsofIvy]

IF the sequence {a_n} converges to a non-zero number, then the sequence {(-1)ⁿa_n} 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, {a_n} converges to 0, {(-1)ⁿa_n} also converges to 0.