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

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The discussion focuses on proving the limit of the function -3x + 1 as x approaches 2, specifically showing that lim (-3x + 1) = -5. The epsilon-delta proof involves demonstrating that for any ε > 0, a corresponding δ can be found such that | -3x + 1 + 5 | < ε whenever 0 < |x - 2| < δ. The calculation shows that | -3x + 6 | simplifies to 3|x - 2|, leading to the conclusion that choosing δ = ε/3 satisfies the condition. This establishes the limit as required, confirming the validity of the epsilon-delta approach.
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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.
 
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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:

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