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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→−3(3x+1)=−8\lim_{x \to -3} \left( 3 x + 1 \right) = -8
Invent a new one:
Femtobytes
Epsilon-delta proof that the limit of 3x + 1 as x approaches -3 is -8, step by step
femtobytes.com/solve/epsilon-delta-limit/0_0_3_1_-3_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→−3(3x+1)=−8\lim_{x \to -3} \left( 3 x + 1 \right) = -8
The result
lim⁡x→−3(3x+1)=−8\lim_{x \to -3} \left( 3 x + 1 \right) = -8

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 = -8L − εc − δc = -3c + 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=−3c = -3, L=−8L = -8, 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=−3x = -3 into the function:

3(−3)+1\displaystyle 3 \left( -3 \right) + 1==−9+1\displaystyle -9 + 1
Do each multiplication. checked
==−8\displaystyle -8
Add the numbers up. checked
Claim: lim⁡x→−3(3x+1)=−8\text{Claim: } \lim_{x \to -3} \left( 3 x + 1 \right) = -8
This is what we will prove, with c=−3c = -3 and L=−8L = -8.
L + εL = -8L − εc − δc = -3c + 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=−3c = -3, L=−8L = -8, 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+1−(−8)∣\displaystyle \left| 3 x + 1 - \left( -8 \right) \right|==∣3x+9∣\displaystyle \left| 3 x + 9 \right|
Combine the plain numbers: 1−(−8)=91 - \left( -8 \right) = 9. checked
==∣3(x−(−3))∣\displaystyle \left| 3 \left( x - \left( -3 \right) \right) \right|
Take out the common factor 33. Check by multiplying back: 3⋅3=93 \cdot 3 = 9. checked
==3∣x−(−3)∣\displaystyle 3 \left| x - \left( -3 \right) \right|
The absolute value of a product is the product of the absolute values, and ∣3∣=3\left| 3 \right| = 3. checked
L + εL = -8L − εc − δc = -3c + 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=−3c = -3, L=−8L = -8, 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