Lim of An=(n+1)^(1/3) - n^(1/3)

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In summary, the conversation discusses solving a limit involving cube roots and the use of the binomial theorem. The limit is found to be 0, and the conversation also mentions other possible methods for solving the limit. The conversation also discusses using a ratio/root test and a Taylor series to find the limit.
  • #1
oferon
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Hi all!
Been trying to look for some examples with no luck.. all I found is
related to square roots, not cube roots..
Anyway I'm trying to solve: [tex]\lim_{n\to\infty}\sqrt[3]{n+1} - \sqrt[3]{n}[/tex]

The limit is obviously 0.. But how do I simplify this expression to show it?
Or should I use the ratio\root test? Even though I couldn't..
Thanks a bunch
 
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  • #2
[tex] (n+x)^{1/3} = n^{1/3}+\frac{x}{3 n^{2/3}}-\frac{x^2}{9 n^{5/3}}+\frac{5 x^3}{81 n^{8/3}} + ... [/tex]
[tex] (n+1)^{1/3} - (n)^{1/3} = \frac{1}{3 n^{2/3}} + O(n^{-5/3}) [/tex]
[tex] \lim_{n \to \infty} \left((n+1)^{1/3} - (n)^{1/3}\right) = 0 [/tex]
 
  • #3
This is the binomal theorem right? Is this the only way to solve this lim?
 
  • #4
I just expanded using a Taylor series. There's probably other ways.
 
  • #5
Yep, the trick is using [tex] a^3-b^3=(a-b)(a^2+ab+b^2)[/tex]

Thanks for your help anyway.
 
  • #6
Not sure how that helps, but if you found some way it does great!
 
  • #7
Oh okay I see. You know it's positive and then you bound it above using what you wrote.
 
  • #8
oferon said:
Yep, the trick is using [tex] a^3-b^3=(a-b)(a^2+ab+b^2)[/tex]

Thanks for your help anyway.

[itex]Let \ a \ = \ \ \sqrt[3]{n + 1} \ \ and \ \ b \ \ = \ \sqrt[3]{n}.[/itex]


[itex]Then \ \ \dfrac{a - b}{1} \cdot \dfrac{a^2 + ab + b^2}{a^2 + ab + b^2} \ =[/itex]


[itex]\dfrac{a^3 - b^3}{a^2 + ab + b^3} [/itex]



What does the numerator become?


As n --> oo, what happens to the denominator?
 

1. What is the limit of the sequence An=(n+1)^(1/3) - n^(1/3)?

The limit of the sequence An=(n+1)^(1/3) - n^(1/3) is 0.

2. How do you find the limit of a sequence?

To find the limit of a sequence, you need to take the limit as n approaches infinity. This means evaluating the expression for larger and larger values of n until you can determine a pattern or trend. If the values get closer and closer to a single number, then that number is the limit.

3. What is the significance of the limit of a sequence?

The limit of a sequence represents the value that the terms in the sequence approach as n becomes infinitely large. It can give insight into the behavior and patterns of the sequence, and can also be used to determine if the sequence converges or diverges.

4. What does it mean if the limit of a sequence is 0?

If the limit of a sequence is 0, it means that the terms in the sequence become infinitely small as n becomes infinitely large. This can indicate that the sequence is converging towards 0, or that it is oscillating between positive and negative values without approaching a specific limit.

5. Can the limit of a sequence be negative?

Yes, the limit of a sequence can be negative. This means that the terms in the sequence are approaching a negative value as n becomes infinitely large. It is also possible for the limit to be a positive or zero value, depending on the behavior of the sequence.

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