Question

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

11. \(\begin{array}{l}\mathop {lim}\limits_{x \to 2} \frac{{{x^2} - 2x}}{{{x^2} - x - 2}}\\x = 2.5,2.1,2.05,2.01,2.005,2.001,\\1.9,1.95,1.99,1.995,1.999\end{array}\)

Short answer

The value of \(\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 2x}}{{{x^2} - x - 2}}\) is \(0.6 = \frac{2}{3}\).

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,

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 2x}}{{{x^2} - x - 2}}\\x = 2.5,2.1,2.05,2.01,2.005,2.001,\\1.9,1.95,1.99,1.995,1.999\end{array}\)

Value of the limit

The value of limit \(\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 2x}}{{{x^2} - x - 2}}\)with\(x = 2.5,2.1,2.05,2.01,2.005,2.001,1.9,1.95,1.99,1.995,1.999\) will be following:

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

X	f(x)
2.5	0.714286
2.1	0.677419
2.05	0.672131
2.01	0.667774
2.005	0.667221
2.001	0.666778
1.9	0.655172
1.95	0.661017
1.99	0.665552
1.995	0.666110
1.999	0.666556

From the above table, it appears that

\(\mathop {\lim }\limits_{x \to 2} \frac{{{x^2} - 2x}}{{{x^2} - x - 2}} = 0.6 = \frac{2}{3}\)