Delta Epsilon Understanding

Question

I have a problem which states:

Let $f(x)=4x-2$ and $\epsilon > 0$.

I must find a $\delta>0$ s.t. $0<|x-1|<\delta$ implies $|f(x)-2|<\epsilon$.

How can I solve a problem such as this?

Answer 1

Two ways. Directly forward, and backwords.

Forward: Just keep asking yourself: If $|x-1| \le \delta$, what does that say about $4x -2$ and $2$.

$|x - 1| \le \delta \implies$

$-\delta < x - 1 < \delta \implies$

$1 - \delta < x < 1 + \delta \implies$

$4 - 4\delta < 4x < 4 + 4\delta \implies$

$2 - 4\delta < 4x - 2 < 2 + 4\delta \implies$

$2 - 4\delta < f(x) < 2 + 4\delta \implies$

$-4\delta < f(x) - 2 < 4\delta \implies $

$|f(x) -2| < 4\delta$.

So if $\epsilon = 4\delta$ or $\delta = \epsilon /4$ then $|x-1| < \delta \implies |f(x) -2 | \epsilon$

Backwards: Keep asking yourself, what would imply $|f(x)-2| < \epsilon$

$|f(x)-2| < \epsilon \Leftarrow$

$-\epsilon < f(x) -2 < \epsilon \Leftarrow$

$2 - \epsilon < f(x) < 2 + \epsilon \Leftarrow$

$2 - \epsilon < 4x - 2 < 2 + \epsilon \Leftarrow$

$4 - \epsilon < 4x < 4 + \epsilon \Leftarrow$

$1 - \epsilon/4 < x < 1 + \epsilon/4 \Leftarrow$

$-\epsilon/4 < x - 1 < \epsilon/4 \Leftarrow$

$|x - 1|< \epsilon/4 \Leftarrow$

$|x-1| < \delta = \epsilon/4$

So if $\delta = \epsilon/4$ then $|x-1| \delta \implies |f(x) -2| < \epsilon$.

Answer 2

Let $\epsilon > 0$ then $|f(x)-2|=|4x-2-2|=|4x-4|=4|x-1|$ select $\delta = \epsilon / 4$, then $|f(x)-2| < \epsilon $ whenever $|x-1| < \delta$.

Answer 3

$f (x)-2=4 (x-1) $.

We look for $\delta>0$ such that

$$|x-1|<\delta\implies 4|x-1|<\epsilon $$

or $$|x-1|<\delta\implies |x-1|<\frac {\epsilon}{4} $$

a $\delta\leq \frac {\epsilon}{4} $ works.