Q. 26 - Calculus

Question

Use algebra to find the largest possible value of δ or smallest possible value of N that makes each implication is true. Then verify and support your answers with labeled graphs.

If $x\in \left(3-\delta ,3\right)$, then $\frac{1}{3-x}>1000$.

Step-by-Step Solution

Answer

The required value of N is $\frac{1}{1000}$.

Step 1. Given information.

Consider the given question,

If $x\in \left(3-\delta ,3\right)$, then $\frac{1}{3-x}>1000$

Step 2. Find the value of δ .

Write the interval notation $x\in \left(3-\delta ,3\right)$ in inequality first,

$3-\delta <x<3$

Subtract $3$ from every side,

$$\begin{gathered} 3-\delta -3<x-3<3-3 \\ -\delta <x-3<0 \\ \delta >x-3>0 \end{gathered}$$

From $\frac{1}{3-x}>1000$, we can write,

$$\begin{gathered} 3-x<\frac{1}{1000} \\ -\left(x-3\right)<-\frac{1}{1000} \\ x-3>\frac{1}{1000} \end{gathered}$$

As $\delta >x-3>0$, then,

$\delta =\frac{1}{1000}$