Determine $\delta$ if $\varepsilon$ $=$ $0.06$ for $\lim_{x \to 1} (2 - 3x)$ $=$ $-1$

Question

Determine $\delta$ if $\varepsilon$ $=$ $0.06$ for $\lim_{x \to 1} (2 - 3x)$ $=$ $-1$

Book solution:

If $0$ $<$ $|x - 1|$ $<$ $\delta$, then $3|1 - x|$ $<$ $\varepsilon$. $|1-x|$ is the same as $|x-1|$, so $|x-1|$ $<$ $\varepsilon$ and $\delta$ is determined by:

$\delta$ $\leq$ $\frac{\varepsilon}{3}$ $\leq$ $\frac{0.06}{3}$ $\leq$ $0.02$

My solution:

If $0$ $<$ $|x - 1|$ $<$ $\delta$, then $3|1 - x|$ $<$ $\varepsilon$. $|1-x|$ is the same as $|x-1|$, so $|x-1|$ $<$ $\varepsilon$ and $\delta$ is determined by:

$\delta$ $\geq$ $\frac{\varepsilon}{3}$ $\geq$ $\frac{0.06}{3}$ $\geq$ $0.02$

Why?

$3|1 - x|$ $<$ $\varepsilon$ (divide both sides by 3)

$|x - 1|$ $<$ $\frac{\varepsilon}{3}$ ($|1-x|$ the same as $|x-1|$)

We also know that

$0$ $<$ $|x - 1|$ $<$ $\frac{\varepsilon}{3}$

Let $\delta$ $=$ $\frac{\varepsilon}{3}$.

The inequality $0$ $<$ $|x - 1|$ $<$ $\delta$ will still be true if $\delta$ $>$ $\frac{\varepsilon}{3}$ or if it's $\delta$ $=$ $\frac{\varepsilon}{3}$. Thus, $\delta$ $\geq$ $\frac{\varepsilon}{3}$

Verification: ???

$0$ $<$ $|x - 1|$ $<$ $\frac{\varepsilon}{3}$

$0$ $<$ $3|x-1|$ $<$ $\varepsilon$

$3|x-1|$ $<$ $\varepsilon$

Answer 1

You aren't trying to prove if $|(2-3x)-(-1)| < \epsilon \implies 0 < |x-1| < \delta$. (That's a trivial statement; we can usually make that statement true by taking very large $\delta$ whether or not our function is continuous. [not always but usually].)

You are trying to prove if $0 < |x-1| < \delta \implies |(2-3x)-(-1)|\epsilon$ and we are trying to find a $\delta$ where that will work for that $\epsilon$.

.....

To show that a $\delta \ge \frac {\epsilon}3$ won't work, lets suppose $\frac {\epsilon}3 \le |x-1| < \delta$.

Then either $\frac {\epsilon}3 \le x-1 < \delta$ or $\frac {\epsilon}3 \le 1-x < \delta$.

So either $\epsilon \le 3(x-1) < 3\delta$ of $\epsilon \le 3(1-x) < 3\delta$.

So either $0 < \epsilon \le 3x -3 < 3x -2$ or $0 < \epsilon \le 3-3x$.

So either $0 < \epsilon \le (3x-2) +(-1)$ or $0 < \epsilon \le (2-3x) +1 = (2-3x)-(-1)$.

EIther way we end up with $\epsilon \le |(2-3x) -(-1)|$ and ... we failed.

......

We need to have $\delta \le \frac \epsilon 3$ in order to force having $|x-1| <\delta$ to always result in $|(2-3x) -(-1)| < \epsilon$.

But as long as $|x-1| < \delta \le \frac \epsilon 3$ we are good.

If $|x-1| < \delta \le \frac \epsilon 3$ then

$|1-x| < \delta \le \frac \epsilon 3$ and so

$-\frac \epsilon 3\le \delta < 1-x < \delta \le \frac \epsilon 3 \implies$

$-\epsilon \le 3\delta < 3-3x = (2-3x) -(-1) < 3\delta\le\epsilon\implies$

$|(2-3x)-(-1)| < 3\delta \le \epsilon$

and we have ... succeeded.

......

For any $\epsilon > 0$ if we pick any $0< \delta \le \frac \epsilon 3$ we will have that if we pick any $x$ so that $|x-1| < \delta$ it will always follow that $|(2-3x) -(-1)| < \epsilon$.

THAT is what we needed to show and picking a $\delta > \frac {\epsilon}3$ would not allow us to show that.

=====

Then how would you get to δ ≤ ε3 in the first place?

Okay... We'd work backwards. We'd want an argument where the last step is

$|(2-3x)-(-1)| < 0.06 =\epsilon$

And the first step is $|x-1| < \delta$

SO the second to last step is: $-\epsilon = -0.06 < (2-3x) -(-1) < 0.06 = \epsilon$

The third to last step is: $-\epsilon=-0.06 < 3-3x < 0.06 = \epsilon$.

the fourth to last step is: $-\frac {\epsilon}3 = -0.02 < 1-x < 0.02 = \frac \epsilon 3$

Our fifth to last step is: $|1-x| < 0.02=\frac \epsilon 3$

Our sixth to last step is: $|x-1| < 0.2 =\frac \epsilon 3= \delta$.

And the sixth to last step is our first step and we have $\delta = \frac \epsilon 3 = 0.2$.

.....

or alternatively we can work forward with a goal in mind.

We start with $|x-1| < \delta$ and we need that to imply $|(2-3x)-(-1)| < \epsilon$.

So we state $-\delta < x-1 < \delta$ and we need that to imply $|(2-3x)-(-1)| < \epsilon$.

So so state $1-\delta < x < 1+\delta$ and we need that to imply $|(2-3x)-(-1)| < \epsilon$.

So we state $3-3\delta < 3x < 3 + 3\delta$ and we need that to imply $|(2-3x)-(-1)| < \epsilon$.

So we state $-3+3\delta > -3x >-3 -3\delta$ or in other words $-3 -\delta < -3x < -3+\delta$ and we need that to imply $|(2-3x)-(-1)| < \epsilon$.

SO we state $-1 - 3\delta < 2-3x < -1+3\delta$ and we need that to imply $|(2-3x)-(-1)| < \epsilon$.

So we state $-3\delta <(2-3x) -(-1) < 3\delta$ and we need that to imply $|(2-3x)-(-1)| < \epsilon$.

So we state $|(2-3x)-(-1)| < 3\delta$ and we need that to imply $|(2-3x)-(-1)| < \epsilon$.

Well, if $3\delta = \epsilon$ it will imply exactly that!

So if $\delta =\frac \epsilon 3$ we have

$|x-1| < \delta = \frac \epsilon 3$ will directly imply that $|(2-3x) -(-1)| < 3\delta =\epsilon$.

Answer 2

If you vary $x$ by $\pm\delta$, it is clear that $2-3x$ will vary by $\mp3\delta$. Hence any $\delta\le\dfrac\epsilon3$ should do.