Calc: Epsilon-Delta Proof of lim (-3x+1)=-5 as x->2

[MarkFL]

Here is the question:

Calc: epsilon/delta proof?

Write out an epsilon/delta proof to show that:

lim (-3x+1)=-5
x->2

I have posted a link there to this topic so the OP may see my work.

 

[MarkFL]

Hello Bill:

We are given to prove:

\(\displaystyle \lim_{x\to2}(-3x+1)=-5\)

For any given $\epsilon>0$, we wish to find a $\delta$ so that:

$|-3x+1+5|<\epsilon$ whenever $0<|x-2|<\delta$

To do this, consider:

\(\displaystyle |-3x+1+5|=|-3x+6|=3|x-2|\)

Thus, to make:

\(\displaystyle 3|x-2|<\epsilon\)

we need only make:

\(\displaystyle 0<|x-2|<\frac{\epsilon}{3}\)

We may then choose:

\(\displaystyle \delta=\frac{\epsilon}{3}\)

Verification:

If \(\displaystyle 0<|x-2|<\frac{\epsilon}{3}\), then \(\displaystyle 3|x-2|<\epsilon\) implies:

\(\displaystyle |3x-6|=|-3x+6|=|(-3x+1)-(-5)|<\epsilon\)