91Ó°ÊÓ

We are given a rational function $R\left(x\right)=\frac{x^3-3x^2+4x-12}{x^4-3x^3+x-3}$.

We need to find the values where the function is not defined and to determine whether there is an asymptote or a hole appears at those points.

The limit of the function at $x=3$is given as

role="math" localid="1647071604060" $\begin{array}{rcl}\underset{x→3}{\lim }\frac{x^3-3x^2+4x-12}{x^4-3x^3+x-3}& =& \underset{x→3}{\lim }\frac{(x-3)(x^2+4)}{(x-3)(x^3+1)}\\ & =& \underset{x→3}{\lim }\frac{x^2+4}{x^3+1}\\ & =& \frac{13}{28}\end{array}$

So the limit of $R\left(x\right)$at $x=3$is $\frac{13}{28}$but the function is not defined at $x=3$. This means that there is a hole and discontinuty at the point $\left(3,\frac{13}{28}\right)$.

The limit at $x=-1$is given as

$\begin{array}{rcl}\underset{x→-1}{\lim }\frac{x^3-3x^2+4x-12}{x^4-3x^3+x-3}& =& \underset{x→-1}{\lim }\frac{(x-3)(x^2+4)}{(x-3)(x^3+1)}\\ & =& \underset{x→-1}{\lim }\frac{x^2+4}{x^3+1}\end{array}$

So it can be seen that as $x=-1$, the function value approaches infinity. From the left hand side the graph approaches negative infinity and from the right hand side the graph approaches positive infinity.

So $R\left(x\right)$is discontinuous at $x=-1$and there is a vertical asymptote at $x=-1$.