Solving Epsilon Delta Proof: lim 3 as x->6 & lim -1 as x->2

[freezer]

Homework Statement

lim 3 as x->6
lim -1 as x->2

Homework Equations

In the first weeks of a calculus class and doing these epsilon delta proofs.

As i am looking at two of the problems i have been assigned:

Lim 3 as x->6
Lim -1 as x->2

The Attempt at a Solution

considering both are horzontal lines, if i give some ε then there is no intersection with the function and thus no δ. ε must be > 0 so not sure how our proof statement will work:

Given ε>0, choose δ=ε. If 0<0<δ, then |0|<δ, then |3-3| = |0| < δ=ε thus |3-3|<ε whenever 0<|0|<δ. Therefore, by the definition of a limit, lim 3 = 3 as x->6

 

[SammyS]

Homework Statement

lim 3 as x->6
lim -1 as x->2

Homework Equations

In the first weeks of a calculus class and doing these epsilon delta proofs.

As i am looking at two of the problems i have been assigned:

Lim 3 as x->6
Lim -1 as x->2

The Attempt at a Solution

considering both are horizontal lines, if i give some ε then there is no intersection with the function and thus no δ. ε must be > 0 so not sure how our proof statement will work:

Given ε>0, choose δ=ε. If 0<0<δ, then |0|<δ, then |3-3| = |0| < δ=ε thus |3-3|<ε whenever 0<|0|<δ. Therefore, by the definition of a limit, lim 3 = 3 as x->6

Hello freezer. Welcome to PF !

One comment first: For a constant function, you can pick anything for δ, often just choose δ = 1.

Now for your proof. What is the general ε - δ formulation for lim_{x → a} f(x) = L ?

For any ε > 0, there exists a δ such that for any x for which 0 < |x - a| < δ, then |f(x) - L| < ε .

You don't have any x the following statement of yours.

... If 0<0<δ, then |0|<δ, then |3-3| = |0| < δ ...​