Determine a delta for an arbitrary epsilon

Question

1. The following graph of the function f satisfies x→3limf(x)=−1.

 Determine a value of epsilon>0

 that satisfies the following: If 0<|x−3|<epsilon, then |f(x)+1|<1.

 

2. Use the precise definition of limit to prove that  

x→2lim(5x+8)=18

Answer 1

1.

The answer to your question is ε = 1.

But you reversed the usual role of ε and δ; see the question 182 below your graph.

2.

I use the definition

x->a lim(f(x)) = k iff

for every ε > 0

there exists δ > 0 so that

for every x satisfying a-δ < x < a+δ

k-ε < f(x) < k+ε

In your case

a = 2

k = 18

f(x) = 5x+8

Assume any given ε > 0.

Set δ = ε/5.

(Note the relationship between the setting of δ and the slope of the function.)

Chose any x satisfying

2-δ < x < 2+δ. (1)

We need to prove

18-ε < f(x) < 18+ε (2)

Multiply inequality (1) by 5

10-5δ < 5x < 10+5δ (3)

Add 8 to (3)

18-5δ < 5x+8 < 18+5δ

The above is the desired identity (2) after the two substitutions

δ = ε/5

5x+8 = f(x)