SOLUTION: Prove that the limit as x approaches 5 of f(x) = ax+b is 5a+b using delta-epsilon proof.

Question 1027615: Prove that the limit as x approaches 5 of f(x) = ax+b is 5a+b using delta-epsilon proof.
Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
We start by determining a value for delta before starting the proof
:
|f(x) - L| < epsilon, where L is the limit
:
|ax+b - 5a-b| < epsilon
:
|a||x-5| < epsilon
:
|x-5| < epsilon/a
:
|x-c| < delta
:
Note that c=5, since x --> 5
:
Therefore
:
delta = epsilon/a
:
*****************
Proof starts
*****************
:
epsilon > 0 is given
:
delta = epsilon/a
:
epsilon > 0, then delta > 0
:
0 < |x-c| < delta implies
:
|x-5| < epsilon/a
:
|ax-5a| < epsilon
:
Therefore
:
limit as x-->5 (ax+b) = (5a+b)
:
Note that we added b to both sides of the -
:

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