# Epsilon Delta Proof

1. Sep 5, 2012

### freezer

1. The problem statement, all variables and given/known data

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

2. Relevant 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

3. 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

2. Sep 5, 2012

### SammyS

Staff Emeritus
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 limx → 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 ...​