help with this problem

[bartmathhpro]

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

 

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[Steven G]

Determine a value of epsilon>0

that satisfies the following: If 0<|x−3|<epsilon, then |f(x)+1|<1.
That is not true! It should say
Determine a value of delta>0

that satisfies the following: If 0<|f(x)−3|<epsilon, then |f(x)+1|<delta.
You need to find out what f(x) equals---at least around x= 3 (say from x=1 to x=4). Hint: the answer is in the graph!

 

[Luis]

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

 

[Luis]

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

 

[Steven G]

Given epsilon you need to find delta.
In the end it should say delta equals, not epsilon equals. delta = epsilon/5, not epsilon = 5delta.

 

[Steven G]

Just by looking at the graph you should see that x+y=2. Then y =-x+2. You method did work. Very good.

So you are saying that no matter what epsilon equals, delta = 1.
Suppose epsilon is very small, say epsilon = .00001, then delta equals 1?
This means that if you compute f(x) values for x between 2 and 4, then f(x) will be between -1-.00001 and -1+.00001. I think not.
In your work you never had epsilon!