# What is the limit of ( x^3 - 8 )/ (x-2) as x approaches 2?

Jun 17, 2015

The limit is $12.$

#### Explanation:

Notice how you have a difference of two cubes.

${x}^{3} - {y}^{3} = {x}^{3} + {x}^{2} y + x {y}^{2} - {x}^{2} y - x {y}^{2} - {y}^{3}$

$= \left(x - y\right) \left({x}^{2} + x y + {y}^{2}\right)$

In our expression, we have $y = 2$:

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

At this point, the limit can be evaluated:

$\implies \textcolor{b l u e}{{\lim}_{x \to 2} \frac{{x}^{3} - 8}{x - 2}}$

$= {\lim}_{x \to 2} {x}^{2} + 2 x + 4$

$= {\left(2\right)}^{2} + 2 \left(2\right) + 4$

$= 4 + 4 + 4$

$= \textcolor{b l u e}{12}$

Sep 24, 2017

$12.$

#### Explanation:

We can have another soln., if we use the following useful Standard Limit :

${\lim}_{x \to a} \frac{{x}^{n} - {a}^{n}}{x - a} = n \cdot {a}^{n - 1} \ldots \ldots \ldots \ldots . . \left(\star\right) .$

Accordingly,

${\lim}_{x \to 2} \frac{{x}^{3} - 8}{x - 2} ,$

$= {\lim}_{x \to 2} \frac{{x}^{3} - {2}^{3}}{x - 2} ,$

$= 3 \cdot {2}^{3 - 1} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . \left[\because , \left(\star\right)\right] ,$

$= 3 \cdot {2}^{2} ,$

$= 12 ,$ as Respected EEt-AP has already obtained!