Q. 49 - 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 $x\in (c-\delta ,c)\cup (c,c+\delta ) \mathrm{then} \mathrm{f}\left(\mathrm{x}\right)\in (\mathrm{L}-\epsilon ,\mathrm{L}+\epsilon )\mathit{ }$.

$\underset{\mathrm{x}\to 0}{\lim }\frac{1-\cos \left(\mathrm{x}\right)}{\mathrm{x}}=0, \mathrm{\epsilon }=\frac{1}{2}$

Step-by-Step Solution

Answer

The largest value of $\delta \approx 1.11$.

Step 1. Given Information.

The given expression is $\underset{\mathrm{x}\to 0}{\lim }\frac{1-\cos \left(\mathrm{x}\right)}{\mathrm{x}}=0, \mathrm{\epsilon }=\frac{1}{2}$.

Step 2. Explanation.

From the given expression, we have, $c=0, L=0$

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

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

$x\in (0-\delta ,0)\cup (0,0+\delta ) \mathrm{then} \frac{1-\cos \left(x\right)}{x}\in (0-\epsilon ,0+\epsilon )\mathit{ }$

Now, the largest value of delta is given by,

$$\begin{gathered} \frac{1-\cos \left(x\right)}{x}=L+\epsilon \\ \frac{1-\cos \left(x\right)}{x}=0+\frac{1}{2} \\ \frac{1-\cos \left(x\right)}{x}=\frac{1}{2} \\ x=2-2\cos \left(x\right) \\ x\approx 1.11 \end{gathered}$$

Thus,

$$\begin{gathered} \delta =1.11-0 \\ \delta \approx 1.11 \end{gathered}$$

Step 3. Conclusion.

The largest value of $\delta \approx 1.11$