Q. 60 - Calculus

Question

Write delta-epsilon proofs for each of the limit statements  in Exercises $47–60.$

$\underset{x\to 2^{+}}{\lim }3\sqrt{2x-4}=0$

Step-by-Step Solution

Answer

Delta-epsilon proof is, 

Whenever $x\in (2,2+\delta )$, we also have $|3\sqrt{2x-4}-0|<ϵ$.

Step 1. Given information

We are given, 

$\underset{x\to 2^{+}}{\lim }3\sqrt{2x-4}=0$

Step 2. Writing the delta-epsilon proofs

The strategy is to write delta-epsilon proofs for the given limit statement. 

Consider that $\in >0$, choose $\delta =\frac{{ϵ}^2}{18}$.

The limit statement $\underset{x\to 2^{+}}{\lim }3\sqrt{2x-4}=0$ means that for all $\in >0$, there exist $\delta >0$ such that if $x\in (2,2+\delta )$, then $|3\sqrt{2x-4}-0|<ϵ$.

So,

$$\begin{gathered} 2<x<2+\delta \\ 2-2<x-2<2+\delta -2 \\ 0<x-2<\delta \end{gathered}$$

This means that, when $0<x-2<\delta$, we have

$$\begin{gathered} |3\sqrt{2x-4}-0|=\left|3\sqrt{2x-4}\right| \\ =3\sqrt{2(x-2)} \\ <\sqrt{18}\sqrt{\delta } \\ =\sqrt{18}\sqrt{\frac{{\in }^2}{18}} \\ =\in \end{gathered}$$

So, whenever $x\in (2,2+\delta )$, we also have $|3\sqrt{2x-4}-0|<ϵ$.