Q. 81 - Calculus

Question

Write a delta–epsilon proof that shows that the function $f\left(x\right)=3x-5$ is continuous at $x=2$. (This exercise depends on Section 1.3.)

Step-by-Step Solution

Answer

Hence we proved $\underset{x\to 2}{\lim }\left(3x-5\right)=1$.

Step 1. Given Information

We are given the function $f\left(x\right)=3x-5$ is continuous at $x=2$ and we need to write a delta–epsilon proof.

Step 2. Delta–epsilon proof

$$\begin{gathered} f\left(x\right)=3x-5 \\ c=2 \\ L=3\left(2\right)-5 \\ =6-5 \\ =1 \end{gathered}$$

For $\epsilon >0, \delta =\frac{\epsilon }{3}$,

If $0<\left|x-2\right|<\delta$ we have,

$$\begin{gathered} \left|\left(3x-5\right)-1\right| \\ =\left|3x-6\right| \\ =\left|3\left(x-2\right)\right| \\ <3\delta \\ =3\left(\frac{\epsilon }{3}\right) \\ =\epsilon \\ \therefore \underset{x\to 2}{\lim }\left(3x-5\right)=1 \end{gathered}$$