# Question c1d5e

Oct 24, 2016

The limit does not exist

#### Explanation:

The limit ${\lim}_{x \to a} f \left(x\right)$ exists if and only if the right and left hand limits exist and are equal, that is, if and only if

${\lim}_{x \to {a}^{+}} f \left(x\right) = {\lim}_{x \to {a}^{-}} f \left(x\right)$

Note that

$| 2 {x}^{3} - {x}^{2} | = | {x}^{2} \left(2 x - 1\right) |$

$= {x}^{2} | 2 x - 1 |$

$= \left\{\begin{matrix}- {x}^{2} \left(2 x - 1\right) \mathmr{if} x \le \frac{1}{2} \\ {x}^{2} \left(2 x - 1\right) \mathmr{if} x \ge \frac{1}{2}\end{matrix}\right.$

With that, let's calculate the two one-sided limits at $\frac{1}{2}$.

As we approach $\frac{1}{2}$ from the left:

${\lim}_{x \to \frac{1}{2} {\text{^-)(2x-1)/|2x^3-x^2| = lim_(x->1/2}}^{-}} \frac{2 x - 1}{- {x}^{2} \left(2 x - 1\right)}$

=lim_(x->1/2""^-)-1/x^2

$= - \frac{1}{\frac{1}{2}} ^ 2$

$= - 4$

As we approach $\frac{1}{2}$ from the right:

${\lim}_{x \to \frac{1}{2} {\text{^+)(2x-1)/|2x^3-x^2| = lim_(x->1/2}}^{+}} \frac{2 x - 1}{{x}^{2} \left(2 x - 1\right)}$

=lim_(x->1/2""^+)1/x^2#

$= \frac{1}{\frac{1}{2}} ^ 2$

$= 4$

As $4 \ne - 4$, the left and right hand limits are not equal, meaning the limit does not exist.