# limit question

• Feb 26th 2009, 11:25 PM
transgalactic
limit question
$
\lim _{x->+\infty} (x+2)^{\frac{2}{3}} -(x-2)^{\frac{2}{3}}
$

i get infinity - infinity
its undefined
??
• Feb 27th 2009, 12:50 AM
Jameson
Quote:

Originally Posted by transgalactic
$
\lim _{x->+\infty} (x+2)^{\frac{2}{3}} -(x-2)^{\frac{2}{3}}
$

i get infinity - infinity
its undefined
??

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 $\frac{(x+2)^{\frac{4}{3}}-(x-2)^{\frac{4}{3}}}{(x+2)^{\frac{2}{3}}+(x-2)^{\frac{2}{3}}}$
• Feb 27th 2009, 04:29 AM
mr fantastic
Quote:

Originally Posted by transgalactic
$
\lim _{x->+\infty} (x+2)^{\frac{2}{3}} -(x-2)^{\frac{2}{3}}
$

i get infinity - infinity
its undefined
??

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

Quote:

Originally Posted by Jameson
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 $\frac{(x+2)^{\frac{4}{3}}-(x-2)^{\frac{4}{3}}}{(x+2)^{\frac{2}{3}}+(x-2)^{\frac{2}{3}}}$

I'm not so sure this works. The numerator has the form $\infty - \infty$ and so you have the same troubles but worse algebra ....
• Feb 27th 2009, 09:37 AM
Krizalid
Write $\underset{x\to \infty }{\mathop{\lim }}\,\big((x+2)^{2/3}+x^{2/3}-(x-2)^{2/3}-x^{2/3}\big),$ then consider $\sqrt[3]{(x+2)^{2}}-\sqrt[3]{x^{2}}$ and use the identity $a^{3}-b^{3}=(a-b)\left( a^{2}+ab+b^{2} \right).$ The rest is algebra.