# Question #a21df

Jan 25, 2018

If $x$ approaches positive infinity, then the limit is $1$.

If $x$ approaches negative infinity, then the limit is $- 1$.

#### Explanation:

First of all, let's try to factor some things out to make this problem easier.

$2 {x}^{2} - 5 x - 3 = \left(2 x + 1\right) \left(x - 3\right)$

So our limit becomes:

${\lim}_{\left\mid x \right\mid \to \infty} \frac{\left\mid x - 3 \right\mid \left(2 x - 3\right)}{\left(x - 3\right) \left(2 x + 1\right)}$

$\left({\lim}_{\left\mid x \right\mid \to \infty} \frac{\left\mid x - 3 \right\mid}{x - 3}\right) \left({\lim}_{\left\mid x \right\mid \to \infty} \frac{2 x - 3}{2 x + 1}\right)$

$\left({\lim}_{\left\mid x \right\mid \to \infty} \frac{\left\mid x - 3 \right\mid}{x - 3}\right) \cdot 1$

The second limit will converge to 1 no matter what, since the $x$ coefficients on the top and bottom are the same. However, there's a problem with the wording of the question that makes the answer more difficult.

The problem says that the absolute value of $x$ goes to $\infty$.

So, $x$ could be approaching either $\infty$ or $- \infty$

If $x$ is approaching $- \infty$, then $\frac{\left\mid x - 3 \right\mid}{x - 3} = - 1$.

If $x$ is approaching $\infty$, then $\frac{\left\mid x - 3 \right\mid}{x - 3} = 1$.

So now we have two cases:

CASE 1

$x$ approaches $\infty$

${\lim}_{\left\mid x \right\mid \to \infty} \frac{\left\mid x - 3 \right\mid}{x - 3} = 1$

Therefore, the entire limit is 1.

CASE 2

$x$ approaches $- \infty$

${\lim}_{\left\mid x \right\mid \to \infty} \frac{\left\mid x - 3 \right\mid}{x - 3} = - 1$

Therefore, the entire limit is -1.