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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
??
\lim _{x->+\infty} (x+2)^{\frac{2}{3}} -(x-2)^{\frac{2}{3}}
[/tex]
i get infinity - infinity
its undefined
??
eek! you're not allowed to do that!i get infinity - infinity
infinity - infinity is one of several indeterminate forms. None of them are ever an answer.[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...[tex]
\lim _{x->+\infty} (x+2)^{\frac{2}{3}} -(x-2)^{\frac{2}{3}}
[/tex]
i get infinity - infinity
its undefined
??
so i get 0*infinityHi transgalactic!
It's easier to read if you write …
[tex]\lim _{x->+\infty} (x+2)^{2/3} -(x-2)^{2/3}[/tex]
eek! you're not allowed to do that!
Hint: write it [tex]\lim _{x->+\infty} x^{2/3}\left((1 + 2/x)^{2/3} -(1 - 2/x)^{2/3}\right)[/tex]
Nooo! … expand the bracket first, then multiply by the x^{2/3}, then take the limit.so i get 0*infinity
Noo that is also not the result! Read my post and tinytim's again(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.
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?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)(a^{2} + ab + b^{2}) = a^{3} - b^{3}.
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.
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.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...
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?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.