# What is lim_(x->2) (x^5-32)/(x-2) ?

Oct 8, 2016

${\lim}_{x \to 2} \frac{{x}^{5} - 32}{x - 2} = 80$

#### Explanation:

Notice that in general:

${a}^{5} - {b}^{5} = \left(a - b\right) \left({a}^{4} + {a}^{3} b + {a}^{2} {b}^{2} + a {b}^{3} + {b}^{4}\right)$

So we find:

${x}^{5} - 32 = {x}^{5} - {2}^{5}$

$\textcolor{w h i t e}{{x}^{5} - 32} = \left(x - 2\right) \left({x}^{4} + 2 {x}^{3} + {2}^{2} {x}^{2} + {2}^{3} x + {2}^{4}\right)$

$\textcolor{w h i t e}{{x}^{5} - 32} = \left(x - 2\right) \left({x}^{4} + 2 {x}^{3} + 4 {x}^{2} + 8 x + 16\right)$

So we can simplify our rational expression to find:

${\lim}_{x \to 2} \frac{{x}^{5} - 32}{x - 2} = {\lim}_{x \to 2} \frac{\left(\textcolor{red}{\cancel{\textcolor{b l a c k}{x - 2}}}\right) \left({x}^{4} + 2 {x}^{3} + 4 {x}^{2} + 8 x + 16\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{x - 2}}}}$

$\textcolor{w h i t e}{{\lim}_{x \to 2} \frac{{x}^{5} - 32}{x - 2}} = {\lim}_{x \to 2} \left({x}^{4} + 2 {x}^{3} + 4 {x}^{2} + 8 x + 16\right)$

$\textcolor{w h i t e}{{\lim}_{x \to 2} \frac{{x}^{5} - 32}{x - 2}} = {\left(\textcolor{b l u e}{2}\right)}^{4} + 2 {\left(\textcolor{b l u e}{2}\right)}^{3} + 4 {\left(\textcolor{b l u e}{2}\right)}^{2} + 8 \left(\textcolor{b l u e}{2}\right) + 16$

$\textcolor{w h i t e}{{\lim}_{x \to 2} \frac{{x}^{5} - 32}{x - 2}} = 80$