# How do you find the limit of abs(6x^2-x-5) / [2x^2+2x-4]  as x approaches 1?

Apr 25, 2016

The two sided limit does not exist. For one sided limits see below.

#### Explanation:

${\lim}_{x \rightarrow 1} \frac{\left\mid 6 {x}^{2} - x - 5 \right\mid}{2 {x}^{2} + 2 x - 4}$ has initial form $\frac{0}{0}$, which is indeterminate.

Since $1$ is a zero of the poynomials in the numerator and the denominator, we can be sure that $x - 1$ is a factor of both.

$\frac{\left\mid 6 {x}^{2} - x - 5 \right\mid}{2 {x}^{2} + 2 x - 4} = \frac{\left\mid 6 x + 5 \right\mid \left\mid x - 1 \right\mid}{2 \left(x + 2\right) \left(x - 1\right)}$

Observe that

$\frac{\left\mid x - 1 \right\mid}{x - 1} = \left\{\begin{matrix}1 & \text{if" & x > 1 \\ -1 & "if} & x < 1\end{matrix}\right.$

So,

$\frac{\left\mid 6 x + 5 \right\mid \left\mid x - 1 \right\mid}{2 \left(x + 2\right) \left(x - 1\right)} = \left\{\begin{matrix}\frac{\left\mid 6 x + 5 \right\mid}{2 x + 4} & \text{if" & x > 1 \\ -abs(6x+5)/(2x+4) & "if} & x < 1\end{matrix}\right.$

Therefore,

${\lim}_{x \rightarrow {1}^{+}} \frac{\left\mid 6 {x}^{2} - x - 5 \right\mid}{2 {x}^{2} + 2 x - 4} = \frac{11}{6}$

and

${\lim}_{x \rightarrow {1}^{-}} \frac{\left\mid 6 {x}^{2} - x - 5 \right\mid}{2 {x}^{2} + 2 x - 4} = - \frac{11}{6}$

Because the one-sided limits are distinct, the limit does not exist.