$\displaystyle
\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 $\displaystyle \frac{(x+2)^{\frac{4}{3}}-(x-2)^{\frac{4}{3}}}{(x+2)^{\frac{2}{3}}+(x-2)^{\frac{2}{3}}}$
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 ....
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.