$\displaystyle

\lim _{x->+\infty} (x+2)^{\frac{2}{3}} -(x-2)^{\frac{2}{3}}

$

i get infinity - infinity

its undefined

??

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- Feb 26th 2009, 10:25 PMtransgalacticlimit question
$\displaystyle

\lim _{x->+\infty} (x+2)^{\frac{2}{3}} -(x-2)^{\frac{2}{3}}

$

i get infinity - infinity

its undefined

?? - Feb 26th 2009, 11:50 PMJameson
You need to get this into the form of a fraction to apply L'Hopitals rule. I think multiplying this by its conjugate divided by itself should work. It should come out to $\displaystyle \frac{(x+2)^{\frac{4}{3}}-(x-2)^{\frac{4}{3}}}{(x+2)^{\frac{2}{3}}+(x-2)^{\frac{2}{3}}}$

- Feb 27th 2009, 03:29 AMmr fantastic
It's not undefined. It's an indeterminate form.

The limit is 0 (and I'm sure you will recognise this answer ....)

One approach might be to simply apply the generalised binomial theorem to each term and simplify. Then take the limit ....

Another approach might be to re-write as $\displaystyle x^{2/3} \left[ \left( 1 + \frac{2}{x}\right)^{2/3} - \left( 1 - \frac{2}{x}\right)^{2/3}\right] = \frac{\left( 1 + \frac{2}{x}\right)^{2/3} - \left( 1 - \frac{2}{x}\right)^{2/3}}{x^{-2/3}}$ so that you have the indeterminant form $\displaystyle \frac{0}{0}$ and then use l'Hopital's Rule.

I'm not so sure this works. The numerator has the form $\displaystyle \infty - \infty$ and so you have the same troubles but worse algebra .... - Feb 27th 2009, 08:37 AMKrizalid
Write $\displaystyle \underset{x\to \infty }{\mathop{\lim }}\,\big((x+2)^{2/3}+x^{2/3}-(x-2)^{2/3}-x^{2/3}\big),$ then consider $\displaystyle \sqrt[3]{(x+2)^{2}}-\sqrt[3]{x^{2}}$ and use the identity $\displaystyle a^{3}-b^{3}=(a-b)\left( a^{2}+ab+b^{2} \right).$ The rest is algebra.