How do you evaluate the limit ((2+x)^3-8)/x as x approaches 0?

Oct 6, 2016

${\lim}_{x \to 0} \frac{{\left(2 + x\right)}^{3} - 8}{x} = 12$

Explanation:

$\frac{{\left(2 + x\right)}^{3} - 8}{x} = \frac{{2}^{3} + 3 \left({2}^{2}\right) x + 3 \left(2\right) {x}^{2} + {x}^{3} - 8}{x}$

$\textcolor{w h i t e}{\frac{{\left(2 + x\right)}^{3} - 8}{x}} = \frac{\textcolor{red}{\cancel{\textcolor{b l a c k}{8}}} + 12 x + 6 {x}^{2} + {x}^{3} - \textcolor{red}{\cancel{\textcolor{b l a c k}{8}}}}{x}$

$\textcolor{w h i t e}{\frac{{\left(2 + x\right)}^{3} - 8}{x}} = \frac{\left(12 + 6 x + {x}^{2}\right) \textcolor{red}{\cancel{\textcolor{b l a c k}{x}}}}{\textcolor{red}{\cancel{\textcolor{b l a c k}{x}}}}$

$\textcolor{w h i t e}{\frac{{\left(2 + x\right)}^{3} - 8}{x}} = 12 + 6 x + {x}^{2}$

with exclusion $x \ne 0$

Hence:

${\lim}_{x \to 0} \frac{{\left(2 + x\right)}^{3} - 8}{x} = {\lim}_{x \to 0} \left(12 + 6 x + {x}^{2}\right) = 12$

Oct 6, 2016

$12$

Explanation:

${a}^{3} - {b}^{3} = \left({a}^{2} + a b + {b}^{2}\right) \left(a - b\right)$

so

$\frac{{\left(2 + x\right)}^{3} - {2}^{3}}{x} = \frac{\left({\left(2 + x\right)}^{2} + \left(2 + x\right) 2 + 4\right) \left(2 + x - 2\right)}{x}$

so

${\lim}_{x \to 0} \frac{{\left(2 + x\right)}^{3} - {2}^{3}}{x} = {\lim}_{x \to 0} \left({\left(2 + x\right)}^{2} + \left(2 + x\right) 2 + 4\right) = 12$