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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→−1(−2x−6)=−4\lim_{x \to -1} \left( -2 x - 6 \right) = -4
Invent a new one:
Femtobytes
Epsilon-delta proof that the limit of -2x - 6 as x approaches -1 is -4, step by step
femtobytes.com/solve/epsilon-delta-limit/0_0_-2_-6_-1_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→−1(−2x−6)=−4\lim_{x \to -1} \left( -2 x - 6 \right) = -4
The result
lim⁡x→−1(−2x−6)=−4\lim_{x \to -1} \left( -2 x - 6 \right) = -4

Proved. Given any tolerance ε>0\varepsilon > 0, the choice δ=ε2\delta = \frac{\varepsilon}{2} 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 = -4L − εc − δc = -1c + 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=−1c = -1, L=−4L = -4, ε=110\varepsilon = \frac{1}{10} and δ=120\delta = \frac{1}{20}. 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=−1x = -1 into the function:

−2(−1)−6\displaystyle -2 \left( -1 \right) - 6==2−6\displaystyle 2 - 6
Do each multiplication. checked
==−4\displaystyle -4
Add the numbers up. checked
Claim: lim⁡x→−1(−2x−6)=−4\text{Claim: } \lim_{x \to -1} \left( -2 x - 6 \right) = -4
This is what we will prove, with c=−1c = -1 and L=−4L = -4.
L + εL = -4L − εc − δc = -1c + 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=−1c = -1, L=−4L = -4, ε=110\varepsilon = \frac{1}{10} and δ=120\delta = \frac{1}{20}. 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.

∣−2x−6−(−4)∣\displaystyle \left| -2 x - 6 - \left( -4 \right) \right|==∣−2x−2∣\displaystyle \left| -2 x - 2 \right|
Combine the plain numbers: −6−(−4)=−2-6 - \left( -4 \right) = -2. checked
==∣−2(x−(−1))∣\displaystyle \left| -2 \left( x - \left( -1 \right) \right) \right|
Take out the common factor −2-2. Check by multiplying back: −2⋅1=−2-2 \cdot 1 = -2. checked
==2∣x−(−1)∣\displaystyle 2 \left| x - \left( -1 \right) \right|
The absolute value of a product is the product of the absolute values, and ∣−2∣=2\left| -2 \right| = 2. checked
L + εL = -4L − εc − δc = -1c + 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=−1c = -1, L=−4L = -4, ε=110\varepsilon = \frac{1}{10} and δ=120\delta = \frac{1}{20}. 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