Q. 59 - Calculus

Question

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

$\underset{x\to 5^{+}}{\lim }\sqrt{x-5}=0$

Step-by-Step Solution

Answer

Delta-epsilon proof is, 

Whenever $x\in (5,5+\delta )$, we also have $|\sqrt{x-5}-0|<\in$.

Step 1. Given information

We are given, 

$\underset{x\to 5^{+}}{\lim }\sqrt{x-5}=0$

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 $\delta ={ϵ}^2$.

The limit statement $\underset{x\to 5^{+}}{\lim }\sqrt{x-5}=0$ means that for all $\in >0$, there exist $\delta >0$ such that if $x\in (5,5+\delta )$, then $|\sqrt{x-5}-0|<ϵ.$

$$\begin{gathered} 5<x<5+\delta \\ 5-5<x-5<5+\delta -5 \\ 0<x-5<\delta \end{gathered}$$

This means that, when $0<x-5<\delta$, we have

$$\begin{gathered} |\sqrt{x-5}-0|=\left|\sqrt{x-5}\right| \\ =\sqrt{x-5} \\ <\sqrt{\delta } \\ =\sqrt{{\in }^2} \\ =\in \end{gathered}$$

So, whenever $x\in (5,5+\delta )$, we also have $|\sqrt{x-5}-0|<\in$.