Q. 43 - 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 $x\in (c-\delta ,c)\cup (c,c+\delta ) \mathrm{then} \mathrm{f}\left(\mathrm{x}\right)\in (\mathrm{L}-\epsilon ,\mathrm{L}+\epsilon )\mathit{ }$.

$\underset{\mathrm{x}\to 2}{\lim }{\mathrm{x}}^3=8, \mathrm{\epsilon }=0.5$

Step-by-Step Solution

Answer

The largest value of $\delta \approx 0.0408$.

Step 1. Given Information.

The given expression is $\underset{\mathrm{x}\to 2}{\lim }{\mathrm{x}}^3=8, \mathrm{\epsilon }=0.5$.

Step 2. Explanation.

From the given expression, we have, $c=2, L=8$.

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

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

$x\in (2-\delta ,2)\cup (2,2+\delta ) \mathrm{Then}\mathit{ }{x}^{\mathit{3}}\in (8-\mathrm{\epsilon },8+\mathrm{\epsilon })$

Now, the largest value of delta is given by,

$$\begin{gathered} x^3=L+\epsilon \\ {x}^3=8+0.5 \\ {x}^3=8.5 \\ x={\left(8.5\right)}^{\frac{1}{3}} \end{gathered}$$

Thus,

$$\begin{gathered} \delta ={\left(8.5\right)}^{\frac{1}{3}}-2 \\ \delta \approx 0.0408 \end{gathered}$$

Step 3. Conclusion.

The largest value of $\delta \approx 0.0408$.