# Math Help - Limit of a function

1. ## Limit of a function

Hi

x-> infinity of x^(1/3) [ (x+8)^(2/3) - (x+7)^(2/3)]

2. $\lim_{n\rightarrow\infty} {(x+8)}^{2/3} - {(x+7)}^{2/3}$ will give you $\infty - \infty$, which is an indeterminate of L'Hôpital's rule. You have to algebraically change it up so as $\lim_{n\rightarrow\infty} \left({(x+8)}^{2/3} - {(x+7)}^{2/3}\right)$ you get $\frac{0}{0}$ or $\frac{\infty}{\infty}$. Finally, take the derivative of the function.

3. Originally Posted by appletree
Hi
Take note that for $x>0\,,\,\,x^{4/3}(A+B)^{4/3}=(Ax+Bx)^{4/3}$ , and also that $A-B=\left(A^{1/3}-B^{1/3}\right)\left(A^{2/3}+A^{1/3}B^{1/3}+B^{1/3}\right)$ :
$x^{1/3} [ (x+8)^{2/3} - (x+7)^{2/3}]=$ $x^{1/3}\,\frac{(x+8)^2-(x-7)^2}{(x+8)^{4/3}+(x+8)^{2/3}(x+7)^{2/3}+(x+7)^{4/3}}=$ $x^{1/3}\,\frac{2x+15}{(x+8)^{4/3}+(x+8)^{2/3}(x+7)^{2/3}+(x+7)^{4/3}}=$
$\frac{2x^{4/3}+15x^{1/3}}{(x+8)^{4/3}+(x+8)^{2/3}(x+7)^{2/3}+(x+7)^{4/3}}\,\frac{1/x^{4/3}}{1/x^{4/3}}=$ $\frac{2+15/x}{\left(1+8/x\right)^{4/3}+\left(1+8/x\right)^{2/3}\left(1+7/x\right)^{2/3}+\left(1+7/x\right)^{4/3}}$ $\xrightarrow [x\to\infty]{}2/3$