Single Variable Calculus, Chapter 2, 2.2, Section 2.2, Problem 33

Find $\displaystyle\lim\limits_{x \rightarrow 1^-} \frac{1}{x^3-1}$ and $\displaystyle \lim\limits_{x \rightarrow 1^+} \frac{1}{x^3-1}$


a.) By evaluating $f(x) = \displaystyle \frac{1}{x^3-1}$ for values of $x$ that approach 1 from the left and from the right.


if
$
\quad \begin{array}{cc}
x^- =& 0.99999 \qquad \displaystyle f(x) =& \frac{1}{(0.99999)^3-1} \qquad &=& -33333.67\\
x^+ =& 1.00001 \qquad \displaystyle f(x) =& \frac{1}{(1.00001)^3-1} \qquad &=& 33333
\end{array}
$



b.) By reasoning:



$\displaystyle \lim\limits_{x \rightarrow 1^-} \frac{1}{x^3-1}$ if $x$ is close to 1 but smaller than 1, the denominator is a very small
negative number. Therefore the quotient is a very large negative number.



$\displaystyle \lim\limits_{x \rightarrow 1^+} \frac{1}{x^3-1}$ if $x$ is close to 1 but larger than, the denominator is a very small
positive number. Therefore the quotient is a very large positive number.




c.) By graphing:







The graph shows that as $x$ approaches 1 from the right, the value of the limit approaches $\infty$. On the other hand, as $x$ approaches
1 from the left, the value of the limit approaches $-\infty$