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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→5(9x−3)=42\lim_{x \to 5} \left( 9 x - 3 \right) = 42
Invent a new one:
Femtobytes
Epsilon-delta proof that the limit of 9x - 3 as x approaches 5 is 42, step by step
femtobytes.com/solve/epsilon-delta-limit/0_0_9_-3_5_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→5(9x−3)=42\lim_{x \to 5} \left( 9 x - 3 \right) = 42
The result
lim⁡x→5(9x−3)=42\lim_{x \to 5} \left( 9 x - 3 \right) = 42

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 = 42L − εc − δc = 5c + 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=5c = 5, L=42L = 42, 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=5x = 5 into the function:

9⋅5−3\displaystyle 9 \cdot 5 - 3==45−3\displaystyle 45 - 3
Do each multiplication. checked
==42\displaystyle 42
Add the numbers up. checked
Claim: lim⁡x→5(9x−3)=42\text{Claim: } \lim_{x \to 5} \left( 9 x - 3 \right) = 42
This is what we will prove, with c=5c = 5 and L=42L = 42.
L + εL = 42L − εc − δc = 5c + 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=5c = 5, L=42L = 42, 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−3−42∣\displaystyle \left| 9 x - 3 - 42 \right|==∣9x−45∣\displaystyle \left| 9 x - 45 \right|
Combine the plain numbers: −3−42=−45-3 - 42 = -45. checked
==∣9(x−5)∣\displaystyle \left| 9 \left( x - 5 \right) \right|
Take out the common factor 99. Check by multiplying back: 9⋅(−5)=−459 \cdot \left( -5 \right) = -45. checked
==9∣x−5∣\displaystyle 9 \left| x - 5 \right|
The absolute value of a product is the product of the absolute values, and ∣9∣=9\left| 9 \right| = 9. checked
L + εL = 42L − εc − δc = 5c + 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=5c = 5, L=42L = 42, 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