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

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

Short answer

The largest value of$\delta =\frac{\mathrm{\pi }}{6}$.

Step-by-step solution

Step 1. Given Information.

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

Step 2. Explanation. 

From the given expression, we have,$c=\mathrm{\pi },\mathrm{L}=\frac{1}{2}$.

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

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

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

Now, the largest value of delta is given by,

$$\begin{gathered} \sin \left(x\right)=L+\epsilon \\ \sin \left(x\right)=0+\frac{1}{2} \\ \sin \left(x\right)=\frac{1}{2} \\ x=\frac{\mathrm{\pi }}{6} \end{gathered}$$

Thus,

$$\begin{gathered} \delta =\frac{\mathrm{\pi }}{6}-0 \\ \delta =\frac{\mathrm{\pi }}{6} \end{gathered}$$

Step 3. Conclusion.

The largest value of$\delta =\frac{\mathrm{\pi }}{6}$.