# How do you find the limit of (x^3 *abs(2x-6))/ (x-3) as x approaches 3?

Oct 2, 2016

The one sided limits disagree, so there is no two sided limit.

#### Explanation:

Note that if $t \ge 0$ then $\left\mid t \right\mid = t$ and if $t \le 0$ then $\left\mid t \right\mid = - t$.

When $x > 3$ then $x - 3 > 0$ and we have:

$\frac{{x}^{3} \cdot \left\mid 2 x - 6 \right\mid}{x - 3} = {x}^{3} \cdot \frac{\left\mid 2 \left(x - 3\right) \right\mid}{x - 3} = \frac{{x}^{3} \cdot 2 \left(\textcolor{red}{\cancel{\textcolor{b l a c k}{x - 3}}}\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{x - 3}}}} = 2 {x}^{3}$

Hence ${\lim}_{x \to {3}^{+}} \frac{{x}^{3} \cdot \left\mid 2 x - 6 \right\mid}{x - 3} = {\lim}_{x \to {3}^{+}} 2 {x}^{3} = 54$

When $x < 3$ then $x - 3 < 0$ and we have:

$\frac{{x}^{3} \cdot \left\mid 2 x - 6 \right\mid}{x - 3} = {x}^{3} \cdot \frac{\left\mid 2 \left(x - 3\right) \right\mid}{x - 3} = \frac{{x}^{3} \cdot \left(- 2 \left(\textcolor{red}{\cancel{\textcolor{b l a c k}{x - 3}}}\right)\right)}{\textcolor{red}{\cancel{\textcolor{b l a c k}{x - 3}}}} = - 2 {x}^{3}$

Hence ${\lim}_{x \to {3}^{-}} \frac{{x}^{3} \cdot \left\mid 2 x - 6 \right\mid}{x - 3} = {\lim}_{x \to {3}^{-}} - 2 {x}^{3} = - 54$

Since the one sided limits disagree there is no two sided limit as $x \to 3$.