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=1100\varepsilon = \frac{1}{100}.

lim⁡x→−2(2x3+2x2+8x+2)=−22\lim_{x \to -2} \left( 2 x^{3} + 2 x^{2} + 8 x + 2 \right) = -22
Invent a new one:
The problem

Prove this limit using only the epsilon-delta definition, then try the proof out with the tolerance ε=1100\varepsilon = \frac{1}{100}.

lim⁡x→−2(2x3+2x2+8x+2)=−22\lim_{x \to -2} \left( 2 x^{3} + 2 x^{2} + 8 x + 2 \right) = -22
The result
lim⁡x→−2(2x3+2x2+8x+2)=−22\lim_{x \to -2} \left( 2 x^{3} + 2 x^{2} + 8 x + 2 \right) = -22

Proved. Given any tolerance ε>0\varepsilon > 0, the choice δ=min⁡{1,  ε36}\delta = \min\left\{ 1, \; \frac{\varepsilon}{36} \right\} works.

checked 18 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 = -22L − εc − δc = -2c + 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=−2c = -2, L=−22L = -22, ε=1100\varepsilon = \frac{1}{100} and δ=13600\delta = \frac{1}{3600}. 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=−2x = -2 into the function:

2(−2)3+2(−2)2+8(−2)+2\displaystyle 2 \left( -2 \right)^{3} + 2 \left( -2 \right)^{2} + 8 \left( -2 \right) + 2==2(−8)+2⋅4+8(−2)+2\displaystyle 2 \left( -8 \right) + 2 \cdot 4 + 8 \left( -2 \right) + 2
Work out each power of −2-2 first. Powers come before multiplying, and multiplying before adding. checked
==−16+8−16+2\displaystyle -16 + 8 - 16 + 2
Do each multiplication. checked
==−22\displaystyle -22
Add the numbers up. checked
Claim: lim⁡x→−2(2x3+2x2+8x+2)=−22\text{Claim: } \lim_{x \to -2} \left( 2 x^{3} + 2 x^{2} + 8 x + 2 \right) = -22
This is what we will prove, with c=−2c = -2 and L=−22L = -22.
2

Rewrite everything in terms of the distance from c

Why this step

We control one thing only: how far xx is from −2-2. But ff is written in powers of xx, which hides that distance. So we give the distance a name, h=x−(−2)h = x - \left( -2 \right), and rewrite the function in powers of hh. After this the quantity we control is the only letter on the page.

From h=x−(−2)h = x - \left( -2 \right) we get x=−2+hx = -2 + h. The condition 0<∣x−c∣<δ0 < \left| x - c \right| < \delta becomes simply 0<∣h∣<δ0 < \left| h \right| < \delta.

2x3+2x2+8x+2−(−22)\displaystyle 2 x^{3} + 2 x^{2} + 8 x + 2 - \left( -22 \right)==2(−2+h)3+2(−2+h)2+8(−2+h)+2−(−22)\displaystyle 2 \left( -2 + h \right)^{3} + 2 \left( -2 + h \right)^{2} + 8 \left( -2 + h \right) + 2 - \left( -22 \right)
Replace every xx by −2+h-2 + h. checked
==2(−8+12h−6h2+h3)+2(4−4h+h2)+8(−2+h)+2−(−22)\displaystyle 2 \left( -8 + 12 h - 6 h^{2} + h^{3} \right) + 2 \left( 4 - 4 h + h^{2} \right) + 8 \left( -2 + h \right) + 2 - \left( -22 \right)
Multiply out the powers: (u+h)2=u2+2uh+h2(u + h)^2 = u^2 + 2uh + h^2 and (u+h)3=u3+3u2h+3uh2+h3(u + h)^3 = u^3 + 3u^2 h + 3u h^2 + h^3, with u=−2u = -2. checked
==2h3−12h2+24h−16+2h2−8h+8+8h−16+2−(−22)\displaystyle 2 h^{3} - 12 h^{2} + 24 h - 16 + 2 h^{2} - 8 h + 8 + 8 h - 16 + 2 - \left( -22 \right)
Multiply each bracket by the number in front of it. A minus sign in front reaches every term inside. checked
==2h3−10h2+24h\displaystyle 2 h^{3} - 10 h^{2} + 24 h
Collect the terms with the same power of hh. The plain numbers add up to exactly 00, and that is no accident: they are f(−2)−Lf(-2) - L, and we CHOSE L=f(−2)L = f(-2). A wrong candidate for LL would leave a number here that no δ\delta could ever shrink. checked
==h(2h2−10h+24)\displaystyle h \left( 2 h^{2} - 10 h + 24 \right)
Every remaining term contains hh, so take hh out as a common factor. checked
∣2x3+2x2+8x+2−(−22)∣=∣h∣⋅∣2h2−10h+24∣\left| 2 x^{3} + 2 x^{2} + 8 x + 2 - \left( -22 \right) \right| = \left| h \right| \cdot \left| 2 h^{2} - 10 h + 24 \right|
Take absolute values. The absolute value of a product is the product of the absolute values. checked
3

Cage the second factor

4

Rough work: choose delta

5

The proof, written forwards

6

Test the rule with a real tolerance