Single Variable Calculus, Chapter 2, 2.2, Section 2.2, Problem 17

Evaluate the function $\displaystyle \lim \limits_{x \to 2} \frac{x^2 - 2x}{x^2 - x - 2}$ at the given numbers
$x = 2.5, 2.1, 2.05, 2.01, 2.005, 2.004, 1.9, 1.95, 1.99, 1.995, 1.999$ 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\\
2.5 & 0.71428 \\
2.1 & 0.67741 \\
2.05 & 0.67213 \\
2.01 & 0.66777 \\
2.005 & 0.66722 \\
2.001 & 0.66677 \\
1.9 & 0.65517 \\
1.95 & 0.66101\\
1.99 & 0.66555\\
1.995 & 0.66611\\
1.999 & 0.66655\\
\hline
\end{array}


\end{aligned}
\end{equation}
$



The values of the given function approaches $0.66$ from both direction, so the value of the function $\displaystyle \lim \limits_{x \to 2}\frac{x^2 -2x}{x^2 - x - 2} = \frac{2}{3}$

$\displaystyle \lim \limits_{x \to 2} \frac{x^2 - 2x}{x^2 - x - 2} = \frac{(2.000001)^2 - (2.000001)}{(2.000001)^2 - 2.000001 - 2} = \frac{2}{3}$