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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(−6x−2)=16\lim_{x \to -3} \left( -6 x - 2 \right) = 16
Invent a new one:
Femtobytes
Epsilon-delta proof that the limit of -6x - 2 as x approaches -3 is 16, step by step
femtobytes.com/solve/epsilon-delta-limit/0_0_-6_-2_-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(−6x−2)=16\lim_{x \to -3} \left( -6 x - 2 \right) = 16
The result
lim⁡x→−3(−6x−2)=16\lim_{x \to -3} \left( -6 x - 2 \right) = 16

Proved. Given any tolerance ε>0\varepsilon > 0, the choice δ=ε6\delta = \frac{\varepsilon}{6} 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 = 16L − 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=16L = 16, ε=110\varepsilon = \frac{1}{10} and δ=160\delta = \frac{1}{60}. 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:

−6(−3)−2\displaystyle -6 \left( -3 \right) - 2==18−2\displaystyle 18 - 2
Do each multiplication. checked
==16\displaystyle 16
Add the numbers up. checked
Claim: lim⁡x→−3(−6x−2)=16\text{Claim: } \lim_{x \to -3} \left( -6 x - 2 \right) = 16
This is what we will prove, with c=−3c = -3 and L=16L = 16.
L + εL = 16L − 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=16L = 16, ε=110\varepsilon = \frac{1}{10} and δ=160\delta = \frac{1}{60}. 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.

∣−6x−2−16∣\displaystyle \left| -6 x - 2 - 16 \right|==∣−6x−18∣\displaystyle \left| -6 x - 18 \right|
Combine the plain numbers: −2−16=−18-2 - 16 = -18. checked
==∣−6(x−(−3))∣\displaystyle \left| -6 \left( x - \left( -3 \right) \right) \right|
Take out the common factor −6-6. Check by multiplying back: −6⋅3=−18-6 \cdot 3 = -18. checked
==6∣x−(−3)∣\displaystyle 6 \left| x - \left( -3 \right) \right|
The absolute value of a product is the product of the absolute values, and ∣−6∣=6\left| -6 \right| = 6. checked
L + εL = 16L − 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=16L = 16, ε=110\varepsilon = \frac{1}{10} and δ=160\delta = \frac{1}{60}. 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