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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(−9x−8)=1\lim_{x \to -1} \left( -9 x - 8 \right) = 1
Invent a new one:
Femtobytes
Epsilon-delta proof that the limit of -9x - 8 as x approaches -1 is 1, step by step
femtobytes.com/solve/epsilon-delta-limit/0_0_-9_-8_-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(−9x−8)=1\lim_{x \to -1} \left( -9 x - 8 \right) = 1
The result
lim⁡x→−1(−9x−8)=1\lim_{x \to -1} \left( -9 x - 8 \right) = 1

Proved. Given any tolerance ε>0\varepsilon > 0, the choice δ=ε9\delta = \frac{\varepsilon}{9} 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 = 1L − 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=1L = 1, ε=110\varepsilon = \frac{1}{10} and δ=190\delta = \frac{1}{90}. 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:

−9(−1)−8\displaystyle -9 \left( -1 \right) - 8==9−8\displaystyle 9 - 8
Do each multiplication. checked
==1\displaystyle 1
Add the numbers up. checked
Claim: lim⁡x→−1(−9x−8)=1\text{Claim: } \lim_{x \to -1} \left( -9 x - 8 \right) = 1
This is what we will prove, with c=−1c = -1 and L=1L = 1.
L + εL = 1L − 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=1L = 1, ε=110\varepsilon = \frac{1}{10} and δ=190\delta = \frac{1}{90}. 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.

∣−9x−8−1∣\displaystyle \left| -9 x - 8 - 1 \right|==∣−9x−9∣\displaystyle \left| -9 x - 9 \right|
Combine the plain numbers: −8−1=−9-8 - 1 = -9. checked
==∣−9(x−(−1))∣\displaystyle \left| -9 \left( x - \left( -1 \right) \right) \right|
Take out the common factor −9-9. Check by multiplying back: −9⋅1=−9-9 \cdot 1 = -9. checked
==9∣x−(−1)∣\displaystyle 9 \left| x - \left( -1 \right) \right|
The absolute value of a product is the product of the absolute values, and ∣−9∣=9\left| -9 \right| = 9. checked
L + εL = 1L − 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=1L = 1, ε=110\varepsilon = \frac{1}{10} and δ=190\delta = \frac{1}{90}. 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