Q. 52 - Calculus

Question

For each limit $\underset{x\to c}{lim}f\left(x\right)=L$in Exercises 43–54, use graphs and algebra to approximate the largest value of $\delta$such that if $x\in (c-\delta ,c)\cup (c,c+\delta ) \mathrm{then} \mathrm{f}\left(\mathrm{x}\right)\in (\mathrm{L}-\epsilon ,\mathrm{L}+\epsilon )\mathit{ }$.

$\underset{\mathrm{x}\to 3}{\lim }\left(2-{\mathrm{x}}^2\right)=-7, \mathrm{\epsilon }=0.001$

Step-by-Step Solution

Answer

The largest value of $\delta =0.0001666$.

Step 1. Given Information.

The given expression is $\underset{\mathrm{x}\to 3}{\lim }\left(2-{\mathrm{x}}^2\right)=-7, \mathrm{\epsilon }=0.001$.

Step 2. Explanation.

From the given expression, we have, $c=3, L=-7$.

The limit expression can be written as a formal statement as below,

For all epsilon positive, there exists a delta positive such that if

$x\in (3-\delta ,3)\cup (3,3+\delta ) \mathrm{then} \left(2-x^2\right)\in (-7-\epsilon ,-7+\epsilon )\mathit{ }$

Now, the largest value of delta is given by, 

$$\begin{gathered} 2-x^2=L+\epsilon \\ 2-x^2=-7-0.001 \\ 2-x^2=-7.001 \\ 2+7.001=x^2 \\ 9.001=x^2 \\ x=\sqrt{9.001} \end{gathered}$$

Thus, 

$$\begin{gathered} \delta =\sqrt{9.001}-3 \\ \delta \approx 0.0001666 \end{gathered}$$

Step 3. Conclusion.

The largest value of $\delta \approx 0.0001666$.