MHB 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:

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

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:

$$|-3x+1+5|=|-3x+6|=3|x-2|$$

Thus, to make:

$$3|x-2|<\epsilon$$

we need only make:

$$0<|x-2|<\frac{\epsilon}{3}$$

We may then choose:

$$\delta=\frac{\epsilon}{3}$$

Verification:

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

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