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February 18, 2026
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Section 2.2 : The Limit
2. For the function \(\displaystyle R\left( t \right) = \frac{{2 - \sqrt {{t^2} + 3} }}{{t + 1}}\) answer each of the following questions.
- Evaluate the function at the following values of \(t\) compute (accurate to at least 8 decimal places).
- -0.5
- -0.9
- -0.99
- -0.999
- -0.9999
- -1.5
- -1.1
- -1.01
- -1.001
- -1.0001
- Use the information from (a) to estimate the value of \(\displaystyle \mathop {\lim }\limits_{t \to \, - 1} \frac{{2 - \sqrt {{t^2} + 3} }}{{t + 1}}\).
Show All Solutions Hide All Solutions
a Evaluate the function at the following values of \(t\) compute (accurate to at least 8 decimal places). Show Solution- -0.5
- -0.9
- -0.99
- -0.999
- -0.9999
- -1.5
- -1.1
- -1.01
- -1.001
- -1.0001
Here is a table of values of the function at the given points accurate to 8 decimal places.
| \(t\) | \(R(t)\) | \(t\) | \(R(t)\) |
|---|---|---|---|
| -0.5 | 0.39444872 | -1.5 | 0.58257569 |
| -0.9 | 0.48077870 | -1.1 | 0.51828453 |
| -0.99 | 0.49812031 | -1.01 | 0.50187032 |
| -0.999 | 0.49981245 | -1.001 | 0.50018745 |
| -0.9999 | 0.49998125 | -1.0001 | 0.50001875 |
b Use the information from (a) to estimate the value of \(\displaystyle \mathop {\lim }\limits_{t \to \, - 1} \frac{{2 - \sqrt {{t^2} + 3} }}{{t + 1}}\). Show Solution
From the table of values above it looks like we can estimate that,
\[\mathop {\lim }\limits_{t \to \, - 1} \frac{{2 - \sqrt {{t^2} + 3} }}{{t + 1}} = \frac{1}{2}\]