Q. 47 - 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 $x\in (c-\delta ,c)\cup (c,c+\delta ) \mathrm{then} \mathrm{f}\left(\mathrm{x}\right)\in (\mathrm{L}-\epsilon ,\mathrm{L}+\epsilon )\mathit{ }$.

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

Step-by-Step Solution

Answer

The largest value of $\delta =\frac{\mathrm{\pi }}{4}$. 

Step 1. Given Information.

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

Step 2. Explanation.

From the given expression, we have, $c=\mathrm{\pi }, \mathrm{L}=0$.

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

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

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

Now, the largest value of delta is given by,

$$\begin{gathered} \sin \left(x\right)=L+\epsilon \\ \sin \left(x\right)=0+\frac{\sqrt{2}}{2} \\ \sin \left(x\right)=\frac{\sqrt{2}}{2} \\ x=\frac{\mathrm{\pi }}{4} \end{gathered}$$

Thus,

$$\begin{gathered} \delta =\frac{\mathrm{\pi }}{4}-0 \\ \delta =\frac{\mathrm{\pi }}{4} \end{gathered}$$

Step 3. Conclusion.

The largest value of $\delta =\frac{\mathrm{\pi }}{4}$.