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

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In summary, the limit of the given expression as n approaches infinity is 0. However, if the question is misstated and actually refers to a series sum, the limit does not exist. Additionally, the limit does not oscillate and the sequence approaches 0 from both the left and right hand sides. The squeeze theorem can be used to find the limit in this case.
  • #1
macjack
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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
 
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  • #2
limit doesn't exist
 
  • #3
ice109 said:
limit doesn't exist

The limit is zero. If the question is misstated and meant to be a series sum then its not.
 
  • #4
Dick said:
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?
 
  • #5
ice109 said:
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:
[tex]\lim_{n \rightarrow \infty} \frac{\sin n}{n} = 0[/tex]
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, [tex]\lim_{n \rightarrow \infty} (-1) ^ n[/tex]. 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.

-------------------------------------------

@macjack: Are you familiar with the Squeeze Theorem?
 
Last edited:
  • #6
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
 
  • #7
macjack said:
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
 
  • #8
thanks ice109

thanks ...
i am a newbie into calculus..didnt face the limits of sequences till now.
anyway thanks for your help.
 
  • #9
now someone help me, why is the limit of the expression 0 and not nonexistant
 
  • #10
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
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
It is zero. Are you doubting it?
 
  • #13
macjack said:
...
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.
 
  • #14
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
As Dick already said, the limit can be defined as follows: a sequence [tex](x_n)_{n \in \mathbb{N}}[/tex] 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 [tex]x_n = 1/n[/tex] has limit zero, because if I give you some number [tex]\epsilon > 0[/tex] you can always give me an n for which [tex]1/n < \epsilon[/tex]. On the other hand, [tex]x_n = \frac{n}{2}[/tex] does not have a limit. For example, if you claim it's "infinity", I can ask you for a number n such that [tex]|\frac{n}{2} - \infty| < \epsilon[/tex] 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
[tex]\lim_{n \to \infty} \frac{n}{2} = \infty[/tex].

Now maybe it's a nice exercise for you to try and prove that way the limit (or its non-existence) of
[tex]\frac{x^n}{x^m}[/tex]
for the cases n < m, n = m and n > m.
 
  • #16
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.
 

1. What is a limit and why is it important in math?

A limit is a fundamental concept in mathematics that represents the value a function approaches as its input approaches a certain value. It is important because it allows us to understand the behavior of a function near a specific point, and is crucial in calculus and other areas of mathematics.

2. How do I find the limit of a function?

To find the limit of a function, you can use various methods such as direct substitution, factoring, or applying algebraic manipulations. If these methods do not work, you can use the properties of limits or graphical analysis to determine the limit.

3. What is the difference between a one-sided limit and a two-sided limit?

A one-sided limit only considers the values of the function as the input approaches the limit from one specific direction, either from the left or the right. A two-sided limit, on the other hand, considers the values of the function from both directions, and the two-sided limit exists only if the one-sided limits from both directions are equal.

4. Can a function have a limit at a point but not be continuous at that point?

Yes, it is possible for a function to have a limit at a point but not be continuous at that point. This can happen if the function has a hole or jump at that point, which means that the function is not defined at that point.

5. How can limits be used in real-life applications?

Limits have many real-life applications, including calculating the maximum and minimum values of a function, determining rates of change, and solving optimization problems. They are also used in physics and engineering to model and predict the behavior of systems.

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