I have to find a number $\delta$ such that if $|x-2|<\delta$ then $|4x-8|<\epsilon$ where $\epsilon=0.1$
Are the steps mentioned below correct?
Step 1: Simplified the expression,
$ 4|x-2|<\epsilon ==> |x-2|<\epsilon/4===> \epsilon=\delta/4 ==> \delta=0.1/4=0.025$
Is this the correct method to approach this problem?
Best Answer
For any $\varepsilon$, you can choose any $\delta\leq \frac{\varepsilon}{4}$. so when $|x-2|<\delta$, then $|4x-8|<4\delta<\varepsilon$