# Question #e382d

Sep 19, 2017

This limit exists for any $a \in \boldsymbol{R}$ except $a = 0$.

#### Explanation:

Let $f \left(x\right) = \frac{P \left(x\right)}{Q \left(x\right)}$ be a rational function, meaning that $P \left(x\right)$ and $Q \left(x\right)$ are polynomials. If the degrees (highest powers) of $P \left(x\right)$ and $Q \left(x\right)$ are the same, then ${\lim}_{x \to \pm \infty} f \left(x\right)$ exists and equals the ratio of the coefficients of the highest powers of $P \left(x\right)$ and $Q \left(x\right)$.

For the example at hand, the numerator expands to

$\left(3 x + 4\right) \left(x - 1\right) \left(2 x + 1\right) = \left(3 {x}^{2} + x - 4\right) \left(2 x + 1\right)$

$= 6 {x}^{3} + 2 {x}^{2} - 8 x + 3 {x}^{2} + x - 4 = 6 {x}^{3} + 5 {x}^{2} - 7 x - 4$.

Therefore,

${\lim}_{x \to \pm \infty} \frac{\left(3 x + 4\right) \left(x - 1\right) \left(2 x + 1\right)}{a {x}^{3} + x - 4} = \frac{6}{a}$ when $a \ne 0$.'

When $a = 0$, then the numerator has a higher degree than the denominator, so the limit does not exist.

When the denominator has a higher degree than the numerator, then the limit is zero.