Answer
$\lim\limits_{x \to 1^{-}}f(x)=-\infty$
$\lim\limits_{x \to 1^{+}}f(x)=\infty$
Work Step by Step
$f(x)=\frac{1}{x^3-1}$
(a) Find the left and right hand limits by evaluating the function.
$\lim\limits_{x \to 1^{-}}\frac{1}{x^3-1}\approx{f(0.99999)=\frac{1}{(0.99999)^3-1}}=-\infty$
$\lim\limits_{x \to 1^{+}}\frac{1}{x^3-1}\approx{f(1.00001)=\frac{1}{(1.00001)^3-1}}=\infty$
(b) By reasoning it out, if $x$ is really close to, but less than 1, $(x^3)$ is close to but less than 1, and $(x^3-1)$ is a very small negative number.
When we divide $1$ by a very small negative number we get a very large negative number.
So, the limit from the left should be a very large negative number, and as we get closer and closer to $x=1$, $f(x)$ approaches $\textbf{negative infinity}$.
Similarly, from the right we get a very large positive number, $\textbf{positive infinity}$.
(c) We can clearly see from the graph that there is an asymptote at $x=1$, and that it approaches $\textbf{negative infinity}$ from the left and $\textbf{positive infinity}$ from the right.
