Section 2.2 : The Limit
1. For the function \(\displaystyle f\left( x \right) = \frac{{8 - {x^3}}}{{{x^2} - 4}}\) answer each of the following questions.
- Evaluate the function at the following values of \(x\) compute (accurate to at least 8 decimal places).
- 2.5
- 2.1
- 2.01
- 2.001
- 2.0001
- 1.5
- 1.9
- 1.99
- 1.999
- 1.9999
- Use the information from (a) to estimate the value of \(\displaystyle \mathop {\lim }\limits_{x \to 2} \frac{{8 - {x^3}}}{{{x^2} - 4}}\).
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a Evaluate the function at the following values of \(x\) compute (accurate to at least 8 decimal places).- 2.5
- 2.1
- 2.01
- 2.001
- 2.0001
- 1.5
- 1.9
- 1.99
- 1.999
- 1.9999
Show Solution
Here is a table of values of the function at the given points accurate to 8 decimal places.
| \(x\) | \(f(x)\) | \(x\) | \(f(x)\) |
|---|---|---|---|
| 2.5 | -3.38888889 | 1.5 | -2.64285714 |
| 2.1 | -3.07560976 | 1.9 | -2.92564103 |
| 2.01 | -3.00750623 | 1.99 | -2.99250627 |
| 2.001 | -3.00075006 | 1.999 | -2.99925006 |
| 2.0001 | -3.00007500 | 1.9999 | -2.99992500 |
b Use the information from (a) to estimate the value of \(\displaystyle \mathop {\lim }\limits_{x \to 2} \frac{{8 - {x^3}}}{{{x^2} - 4}}\). Show Solution
From the table of values above it looks like we can estimate that,
\[\mathop {\lim }\limits_{x \to 2} \frac{{8 - {x^3}}}{{{x^2} - 4}} = - 3\]