# What is lim_(xrarr4) ( x^4 -256) / (x^3 - 64)?

Oct 27, 2015

${\lim}_{x \rightarrow 4} \frac{{x}^{4} - 256}{{x}^{3} - 64} = \frac{16}{3}$

#### Explanation:

Either using polynomial long division or synthetic division we can discover:
$\textcolor{w h i t e}{\text{XXX}} \left({x}^{4} - 256\right) = \left(x - 4\right) \left({x}^{3} + 4 {x}^{2} + 16 x + 64\right)$
and
$\textcolor{w h i t e}{\text{XXX}} \left({x}^{3} - 64\right) = \left(x - 4\right) \left({x}^{2} + 4 x + 16\right)$

As long as $x \ne 4$
$\textcolor{w h i t e}{\text{XXX}} \frac{{x}^{4} - 256}{{x}^{3} - 64} = \frac{{x}^{3} + 4 {x}^{2} + 16 x + 64}{{x}^{2} + 4 x + 16}$

${\lim}_{x \rightarrow 4} \frac{{x}^{4} - 256}{{x}^{3} - 64} = \frac{{\left(4\right)}^{3} + 4 {\left(4\right)}^{2} + 16 \left(4\right) + 64}{{\left(4\right)}^{2} + 4 \left(4\right) + 16}$

$\textcolor{w h i t e}{\text{XXXXXXXXX}} = \frac{4 \left(64\right)}{3 \left(16\right)} = \frac{16}{3}$

Oct 27, 2015

$\frac{16}{3}$

#### Explanation:

lim_(x→4) (x^4 - 256)/(x^3-64)

=lim_(x→4) (x^4 - 4^4)/(x^3-4^3)

i'm multiplying both numerator and denominator by $\left(x - 4\right)$.
the reason for why i came up with this idea is that there is a theorem in limits as follows,
lim_(x→a) (x^n - a^n)/(x-a) = n a^(n-1)

so,
=lim_(x→4) (x^4 - 4^4)/(x-4)*(x-4)/(x^3-4^3)

=lim_(x→4) (x^4 - 4^4)/(x-4)*lim_(x→4) (x-4)/(x^3-4^3)

$= \frac{4 \cdot {4}^{3}}{3 \cdot {4}^{2}}$

$= {4}^{2} / 3$

$= \frac{16}{3}$

Oct 27, 2015

$\frac{16}{3}$