Single Variable Calculus, Chapter 2, 2.2, Section 2.2, Problem 18

Evaluate the function $\displaystyle \lim \limits_{x \to -1} \frac{x^2 - 2x}{x^2 - x - 2}$ at the given numbers
$x = 0, -0.5, -0.9, -0.95, -0.99, -2, -1.5, -1.1, -1.01, -1.001$ and guess the value of the limit, if it exists.


Substituting all the given values of $x$


$
\begin{equation}
\begin{aligned}

\begin{array}{|c|c|}
\hline\\
x & f(x) \\
\hline\\
0 & 0 \\
-0.5 & -1 \\
-0.9 & -9 \\
-0.95 & -19 \\
-0.99 & -99 \\
-0.999 & -999 \\
-2 & 2 \\
-1.5 & 3 \\
-1.1 & 11 \\
-1.01 & 101 \\
-1.001 & 1001\\
\hline
\end{array}


\end{aligned}
\end{equation}
$



Based from the values in the table, we can conclude that the limit of the function does not exist
because of its difference between its values as $x$ approaches -1 from left and right.


$
\begin{equation}
\begin{aligned}
\displaystyle \lim \limits_{x \to -1} \frac{x^2 - 2x}{x^2 - x - 2} =& \frac{(-0.999999)^2 - 2 (-0.999999)}{(-0.999999)^2 - (-0.999999)-2} = -999999\\

\displaystyle \lim \limits_{x \to -1} \frac{x^2 - 2x}{x^2 - x - 2} =& \frac{(-1.000001)^2 - 2 (-1.000001)}{(-1.000001)^2 - (-1.000001)-2} = 1000001

\end{aligned}
\end{equation}
$