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Problem type

Limit of a function from the epsilon-delta definition

What it solves
Prove that a polynomial (a straight line, a quadratic or a cubic) has the limit you expect at a point, using only the epsilon-delta definition: for every output tolerance epsilon, find an input tolerance delta that guarantees it.
Subject
Real Analysis
Topic
Limits from the definition
Also called
epsilon delta, limit of a function, continuity from the definition
Your problem

Prove this limit using only the epsilon-delta definition, then try the proof out with the tolerance ε=110\varepsilon = \frac{1}{10}.

lim⁡x→4(3x−3)=9\lim_{x \to 4} \left( 3 x - 3 \right) = 9
Invent a new one:
Femtobytes
Epsilon-delta proof that the limit of 3x - 3 as x approaches 4 is 9, step by step
femtobytes.com/solve/epsilon-delta-limit/0_0_3_-3_4_1~10
The problem

Prove this limit using only the epsilon-delta definition, then try the proof out with the tolerance ε=110\varepsilon = \frac{1}{10}.

lim⁡x→4(3x−3)=9\lim_{x \to 4} \left( 3 x - 3 \right) = 9
The result
lim⁡x→4(3x−3)=9\lim_{x \to 4} \left( 3 x - 3 \right) = 9

Proved. Given any tolerance ε>0\varepsilon > 0, the choice δ=ε3\delta = \frac{\varepsilon}{3} works.

checked 9 lines of this solution were evaluated with exact arithmetic before it was shown to you. A solution with a false line is never served.
Picture 1 of 2Step 1: Find the candidate limit, then say exactly what must be proved
L + εL = 9L − εc − δc = 4c + δinput xoutput f(x)
The graph of the function near the point. The upright strip is the input window of half-width delta; the level band is the output tolerance of half-width epsilon. Inside the strip the graph never leaves the band. Here c=4c = 4, L=9L = 9, ε=110\varepsilon = \frac{1}{10} and δ=130\delta = \frac{1}{30}. The bright part of the graph is the part inside the strip. The hollow circle marks the point itself, which the definition leaves out.
1

Find the candidate limit, then say exactly what must be proved

Why this step

The definition can only CHECK a number we propose; it never hands us one. For a polynomial the natural candidate is the value at the point itself, because a polynomial is built from nothing but adding and multiplying, and neither of those can make a sudden jump. That is a belief. The proof turns it into a fact.

Put x=4x = 4 into the function:

3⋅4−3\displaystyle 3 \cdot 4 - 3==12−3\displaystyle 12 - 3
Do each multiplication. checked
==9\displaystyle 9
Add the numbers up. checked
Claim: lim⁡x→4(3x−3)=9\text{Claim: } \lim_{x \to 4} \left( 3 x - 3 \right) = 9
This is what we will prove, with c=4c = 4 and L=9L = 9.
L + εL = 9L − εc − δc = 4c + δinput xoutput f(x)
The graph of the function near the point. The upright strip is the input window of half-width delta; the level band is the output tolerance of half-width epsilon. Inside the strip the graph never leaves the band. Here c=4c = 4, L=9L = 9, ε=110\varepsilon = \frac{1}{10} and δ=130\delta = \frac{1}{30}. The bright part of the graph is the part inside the strip. The hollow circle marks the point itself, which the definition leaves out.
2

Measure the distance between an output and the limit

Why this step

The definition is about one quantity only, ∣f(x)−L∣\left| f(x) - L \right|. We do not control it directly; we only control how close xx is to cc. So we rewrite the output distance until the input distance ∣x−c∣\left| x - c \right| appears inside it. Then controlling one controls the other.

∣3x−3−9∣\displaystyle \left| 3 x - 3 - 9 \right|==∣3x−12∣\displaystyle \left| 3 x - 12 \right|
Combine the plain numbers: −3−9=−12-3 - 9 = -12. checked
==∣3(x−4)∣\displaystyle \left| 3 \left( x - 4 \right) \right|
Take out the common factor 33. Check by multiplying back: 3⋅(−4)=−123 \cdot \left( -4 \right) = -12. checked
==3∣x−4∣\displaystyle 3 \left| x - 4 \right|
The absolute value of a product is the product of the absolute values, and ∣3∣=3\left| 3 \right| = 3. checked
L + εL = 9L − εc − δc = 4c + δinput xoutput f(x)
The graph of the function near the point. The upright strip is the input window of half-width delta; the level band is the output tolerance of half-width epsilon. Inside the strip the graph never leaves the band. Here c=4c = 4, L=9L = 9, ε=110\varepsilon = \frac{1}{10} and δ=130\delta = \frac{1}{30}. The bright part of the graph is the part inside the strip. The hollow circle marks the point itself, which the definition leaves out.
3

Rough work: how close must x be kept?

4

The proof, written forwards

5

Test the rule with a real tolerance