Single Variable Calculus, Chapter 2, Review Exercises, Section Review Exercises, Problem 6

Determine $\displaystyle \lim \limits_{x \to 1^+} \frac{x^2 - 9}{x^2 + 2x - 3}$


$
\begin{equation}
\begin{aligned}



\lim \limits_{x \to 1^+} \frac{x^2 - 9 }{x^2 + 2x - 3} &= \lim \limits_{x \to 1^+} \frac{\cancel{(x + 3)} (x - 3)}{\cancel{(x + 3)} (x - 1)}
&& \text{Get the factor of numerator and denominator and cancel out like terms}\\
\\
& = \frac{1 - 3}{1 - 1} = \frac{-2}{0}
&& \text{By substituting value of $ x$ which is 1, the denominator will be zero. So the limit is undefined.}

\end{aligned}
\end{equation}
$


So we set values of $x$ as $x$ approaches 1 from the right

$
\begin{array}{|c|c|}
\hline\\
x & f(x) \\
\hline\\
1.01 & -199 \\
1.001 & -1999 \\
1.0001 & -19999 \\
1.00001 & -199999\\
\hline
\end{array}
$

Referring to the table, the values of $y$ became more negatively large number as $x$ approaches 1 from the right.

Therefore,

$\fbox{$ \lim \limits_{x \to 1^+} \displaystyle \frac{x^2 - 9 }{x^2 + 2x - 3} = -\infty$}$