Question

Guess the value of the limit (if it exists) by evaluating the function at the given numbers (correct to six decimal places).

\(\begin{array}{l}\mathop {lim}\limits_{x \to - 1} \frac{{{x^2} - 2x}}{{{x^2} - x - 2}}\\x = 0. - 0.5. - 0.95, - 0.99, - 0.999,\\ - 2, - 1.5, - 1.1, - 1.01, - 1.001\end{array}\)

Short answer

The \(\mathop {lim}\limits_{x \to - 1} \frac{{{x^2} - 2x}}{{{x^2} - x - 2}}\) does not exist because both the values for x are different.

Step-by-step solution

Introduction

The function does not exist on the given limit but it can sustain on the nearby values of the limit. To find the value of the limit we have to find the nearby limit values.

Given information

The given data is,

\(\begin{array}{l}\mathop {lim}\limits_{x \to - 1} \frac{{{x^2} - 2x}}{{{x^2} - x - 2}}\\x = 0. - 0.5. - 0.95, - 0.99, - 0.999,\\ - 2, - 1.5, - 1.1, - 1.01, - 1.001\end{array}\)

Explanation

Let the given function is, \(f\left( x \right) = \mathop {lim}\limits_{x \to - 1} \frac{{{x^2} - 2x}}{{{x^2} - x - 2}}\).

The value chart of \(f\left( x \right) = \mathop {lim}\limits_{x \to - 1} \frac{{{x^2} - 2x}}{{{x^2} - x - 2}}\) with \(x = 0. - 0.5. - 0.95, - 0.99, - 0.999, - 2, - 1.5, - 1.1, - 1.01, - 1.001\) will be as following,

x	f(x)	x	f(x)
0	-0.5	-2	2
-0.5	-1	-1.5	3
-0.95	-19	-1.1	11
-0.99	-99	-1.01	101
-0.999	-999	-1.001	1001

For,

\(\begin{array}{l}\mathop {lim}\limits_{x \to - 1} \frac{{{x^2} - 2x}}{{{x^2} - x - 2}}\\x = 0. - 0.5. - 0.95, - 0.99, - 0.999,\end{array}\)

Values are going to\( - \infty \)

And for,

\(\begin{array}{l}\mathop {lim}\limits_{x \to - 1} \frac{{{x^2} - 2x}}{{{x^2} - x - 2}}\\x = - 2, - 1.5, - 1.1, - 1.01, - 1.001\end{array}\)

Values are going to\( + \infty \)

Both the values are different, hence \(\mathop {lim}\limits_{x \to - 1} \frac{{{x^2} - 2x}}{{{x^2} - x - 2}}\) does not exist.