Undefined Limit of (x+2)^(2/3) - (x-2)^(2/3) as x Approaches Infinity

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In summary, the limit of the expression \lim _{x->+\infty} (x+2)^{\frac{2}{3}} -(x-2)^{\frac{2}{3}} is undefined, as infinity minus infinity is one of several indeterminate forms. However, when considering the behavior of the function as x approaches infinity, it is clear that the limit is 0, as the difference between (x+2)^{\frac{2}{3}} and (x-2)^{\frac{2}{3}} becomes negligible compared to x^{\frac{2}{3}}. Different approaches, such as expanding the brackets and multiplying by a factor, can also be used to show the limit is 0
  • #1
transgalactic
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[tex]
\lim _{x->+\infty} (x+2)^{\frac{2}{3}} -(x-2)^{\frac{2}{3}}
[/tex]

i get infinity - infinity
its undefined
??
 
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  • #2
Hi transgalactic! :smile:

It's easier to read if you write …

[tex]\lim _{x->+\infty} (x+2)^{2/3} -(x-2)^{2/3}[/tex] :wink:
transgalactic said:
i get infinity - infinity

eek! you're not allowed to do that! :eek:

Hint: write it [tex]\lim _{x->+\infty} x^{2/3}\left((1 + 2/x)^{2/3} -(1 - 2/x)^{2/3}\right)[/tex] :wink:
 
  • #3
transgalactic said:
[tex]
\lim _{x->+\infty} (x+2)^{\frac{2}{3}} -(x-2)^{\frac{2}{3}}
[/tex]

i get infinity - infinity
its undefined
??
infinity - infinity is one of several indeterminate forms. None of them are ever an answer.
 
  • #4
transgalactic said:
[tex]
\lim _{x->+\infty} (x+2)^{\frac{2}{3}} -(x-2)^{\frac{2}{3}}
[/tex]

i get infinity - infinity
its undefined
??

A hint when doing limits, just think about what happens when the number gets REALLY big. Eventually does it matter that you add 2 or subtract 2 from say...10000000000000000? It will approximate to [tex]x^{\frac{2}{3}}-x^{\frac{2}{3}}[/tex] in the end...

Also the result is the same as tinytim's =]
 
  • #5
tiny-tim said:
Hi transgalactic! :smile:

It's easier to read if you write …

[tex]\lim _{x->+\infty} (x+2)^{2/3} -(x-2)^{2/3}[/tex] :wink:


eek! you're not allowed to do that! :eek:

Hint: write it [tex]\lim _{x->+\infty} x^{2/3}\left((1 + 2/x)^{2/3} -(1 - 2/x)^{2/3}\right)[/tex] :wink:

so i get 0*infinity

its not defined either
 
  • #6
transgalactic said:
so i get 0*infinity

Nooo! :cry: … expand the bracket first, then multiply by the x2/3, then take the limit.
 
  • #7
(x-2)2/3 is less than (x+2)2/3 so each time the x gets larger a larger sum is in (x-2) than(x+2) so it goes down... to minus infinity.
 
  • #8
cleopatra said:
(x-2)2/3 is less than (x+2)2/3 so each time the x gets larger a larger sum is in (x-2) than(x+2) so it goes down... to minus infinity.

Noo that is also not the result! Read my post and tinytim's again

When in doubt graph it!
 
  • #9
When tiny-tim says "expand the brackets first" I believe he is referring to the generalized binomial expansion. And you should only need the first few terms.
 
  • #10
Another approach is to multiply the original expression by 1 like so:
[tex](x + 2)^{2/3} - (x - 2)^{2/3} \frac{(x + 2)^{4/3} + (x + 2)^{2/3}(x - 2)^{2/3} + (x - 2)^{4/3}} {(x + 2)^{4/3} + (x + 2)^{2/3}(x - 2)^{2/3} + (x - 2)^{4/3}}[/tex]

This looks pretty complicated, but it's based on a relatively simple idea: that (a - b)(a2 + ab + b2) = a3 - b3.

After using this technique, the numerator can be simplified to a single term, and the limit of the whole expression can be found without much trouble.
 
  • #11
Mark44 said:
Another approach is to multiply the original expression by 1 like so:
[tex](x + 2)^{2/3} - (x - 2)^{2/3} \frac{(x + 2)^{4/3} + (x + 2)^{2/3}(x - 2)^{2/3} + (x - 2)^{4/3}} {(x + 2)^{4/3} + (x + 2)^{2/3}(x - 2)^{2/3} + (x - 2)^{4/3}}[/tex]

This looks pretty complicated, but it's based on a relatively simple idea: that (a - b)(a2 + ab + b2) = a3 - b3.

After using this technique, the numerator can be simplified to a single term, and the limit of the whole expression can be found without much trouble.

Why need to get so complicated :S this is quite a simple problem. Quite frankly the answer is 0. There is an asymptote in the graph tending towards 0 as x increases towards positive infinity. Now which result out of mine or tinytim's did not explain this?

I can re iterate of course "Eventually does it matter that you add 2 or subtract 2 from say...10000000000000000? It will approximate to [tex] x^{\frac{2}{3}}-x^{\frac{2}{3}}[/tex] in the end..." which of course is always tending towards 0...
124h3pt.jpg
 
  • #12
djeitnstine said:
Why need to get so complicated :S this is quite a simple problem. Quite frankly the answer is 0. There is an asymptote in the graph tending towards 0 as x increases towards positive infinity. Now which result out of mine or tinytim's did not explain this?
Agreed that this is a simple problem, and that the limit is 0. Tiny Tim's approach is superior to mine in its simplicity. Your approach, using a computer-generated graph, is useful for gaining understanding about the behavior of this function, but that's about it. It is well-known that computers don't produce exact results, because of the limited precision they use in their calculations, and especially so in raising to fractional powers. And in some cases, they can produce results that are incorrect (e.g., the earlier Intel Pentium math coprocessors back in 1992 or so). You also said that for large x, x + 2 was about the same as x - 2. This is correct, and is helpful in understanding, but it should be said in mathematical terms. Otherwise, it's akin to arm-waving, IMO.

My approach is different from, and slightly more complicated, than Tiny Tim's. But so what? Multiplying by a factor to produce a difference of cubes isn't all that complicated, and the intent was to show another method of approach that might be useful in other problems.

The more tools you have to work with, the more problems you're equipped to solve.
djeitnstine said:
I can re iterate of course "Eventually does it matter that you add 2 or subtract 2 from say...10000000000000000? It will approximate to [tex] x^{\frac{2}{3}}-x^{\frac{2}{3}}[/tex] in the end..." which of course is always tending towards 0...
124h3pt.jpg
 
  • #13
Mark44 said:
Agreed that this is a simple problem, and that the limit is 0. Tiny Tim's approach is superior to mine in its simplicity. Your approach, using a computer-generated graph, is useful for gaining understanding about the behavior of this function, but that's about it. It is well-known that computers don't produce exact results, because of the limited precision they use in their calculations, and especially so in raising to fractional powers. And in some cases, they can produce results that are incorrect (e.g., the earlier Intel Pentium math coprocessors back in 1992 or so). You also said that for large x, x + 2 was about the same as x - 2. This is correct, and is helpful in understanding, but it should be said in mathematical terms. Otherwise, it's akin to arm-waving, IMO.

My approach is different from, and slightly more complicated, than Tiny Tim's. But so what? Multiplying by a factor to produce a difference of cubes isn't all that complicated, and the intent was to show another method of approach that might be useful in other problems.

The more tools you have to work with, the more problems you're equipped to solve.

Well I do understand that your method was valid, however I believe when helping someone whom is slightly confused the simplest approached should be used. Don't you agree?
 
  • #14
I agree that the simplest approach should be shown first. Then you can show other approaches, just as was done in this thread.
 

1. What is the definition of an undefined limit?

An undefined limit is a limit that does not exist or cannot be determined using standard mathematical methods. It may occur when the function approaches different values from the left and right sides, or when the function approaches infinity or negative infinity.

2. How do I determine if a limit is undefined?

To determine if a limit is undefined, you can graph the function and look for any breaks or discontinuities. You can also use algebraic methods, such as factoring or simplifying, to see if the limit can be evaluated.

3. What is the limit of (x+2)^(2/3) - (x-2)^(2/3) as x approaches infinity?

The limit of (x+2)^(2/3) - (x-2)^(2/3) as x approaches infinity is undefined. This is because the function approaches different values from the left and right sides, with no clear limit.

4. Can the undefined limit be evaluated using L'Hopital's rule?

No, L'Hopital's rule only applies to limits that result in indeterminate forms, such as 0/0 or infinity/infinity. Since the limit of (x+2)^(2/3) - (x-2)^(2/3) as x approaches infinity is undefined, L'Hopital's rule cannot be used.

5. Is there a way to rewrite the function to make the limit defined?

No, the function (x+2)^(2/3) - (x-2)^(2/3) cannot be rewritten in a way that would make the limit defined. This is because the function approaches different values from the left and right sides, and there is no way to reconcile these two values to get a single limit.

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