Single Variable Calculus

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

Expert Answers

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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} $

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