# Limit of a function

#### appletree

Hi

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

#### lilaziz1

$$\displaystyle \lim_{n\rightarrow\infty} {(x+8)}^{2/3} - {(x+7)}^{2/3}$$ 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.

#### tonio

Hi

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

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

HallsofIvy