Femtobytes
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(2x−9)=−17\lim_{x \to -4} \left( 2 x - 9 \right) = -17
Invent a new one:
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(2x−9)=−17\lim_{x \to -4} \left( 2 x - 9 \right) = -17
The result
lim⁡x→−4(2x−9)=−17\lim_{x \to -4} \left( 2 x - 9 \right) = -17

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 = -17L − 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=−17L = -17, 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=−4x = -4 into the function:

2(−4)−9\displaystyle 2 \left( -4 \right) - 9==−8−9\displaystyle -8 - 9
Do each multiplication. checked
==−17\displaystyle -17
Add the numbers up. checked
Claim: lim⁡x→−4(2x−9)=−17\text{Claim: } \lim_{x \to -4} \left( 2 x - 9 \right) = -17
This is what we will prove, with c=−4c = -4 and L=−17L = -17.
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−9−(−17)∣\displaystyle \left| 2 x - 9 - \left( -17 \right) \right|==∣2x+8∣\displaystyle \left| 2 x + 8 \right|
Combine the plain numbers: −9−(−17)=8-9 - \left( -17 \right) = 8. checked
==∣2(x−(−4))∣\displaystyle \left| 2 \left( x - \left( -4 \right) \right) \right|
Take out the common factor 22. Check by multiplying back: 2⋅4=82 \cdot 4 = 8. checked
==2∣x−(−4)∣\displaystyle 2 \left| x - \left( -4 \right) \right|
The absolute value of a product is the product of the absolute values, and ∣2∣=2\left| 2 \right| = 2. checked
3

Rough work: how close must x be kept?

4

The proof, written forwards

5

Test the rule with a real tolerance