Hi
Could someone please help me with the general method of finding limits like this one:
x-> infinity of x^(1/3) [ (x+8)^(2/3) - (x+7)^(2/3)]
$\displaystyle \lim_{n\rightarrow\infty} {(x+8)}^{2/3} - {(x+7)}^{2/3} $ will give you $\displaystyle \infty - \infty $, which is an indeterminate of L'Hôpital's rule. You have to algebraically change it up so as $\displaystyle \lim_{n\rightarrow\infty} \left({(x+8)}^{2/3} - {(x+7)}^{2/3}\right) $ you get $\displaystyle \frac{0}{0} $ or $\displaystyle \frac{\infty}{\infty} $. Finally, take the derivative of the function.
Take note that for $\displaystyle x>0\,,\,\,x^{4/3}(A+B)^{4/3}=(Ax+Bx)^{4/3}$ , and also that $\displaystyle A-B=\left(A^{1/3}-B^{1/3}\right)\left(A^{2/3}+A^{1/3}B^{1/3}+B^{1/3}\right)$ :
$\displaystyle x^{1/3} [ (x+8)^{2/3} - (x+7)^{2/3}]=$ $\displaystyle x^{1/3}\,\frac{(x+8)^2-(x-7)^2}{(x+8)^{4/3}+(x+8)^{2/3}(x+7)^{2/3}+(x+7)^{4/3}}=$ $\displaystyle x^{1/3}\,\frac{2x+15}{(x+8)^{4/3}+(x+8)^{2/3}(x+7)^{2/3}+(x+7)^{4/3}}=$
$\displaystyle \frac{2x^{4/3}+15x^{1/3}}{(x+8)^{4/3}+(x+8)^{2/3}(x+7)^{2/3}+(x+7)^{4/3}}\,\frac{1/x^{4/3}}{1/x^{4/3}}=$ $\displaystyle \frac{2+15/x}{\left(1+8/x\right)^{4/3}+\left(1+8/x\right)^{2/3}\left(1+7/x\right)^{2/3}+\left(1+7/x\right)^{4/3}}$ $\displaystyle \xrightarrow [x\to\infty]{}2/3$
Tonio