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→−3(−2x2−3)=−21\lim_{x \to -3} \left( -2 x^{2} - 3 \right) = -21
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→−3(−2x2−3)=−21\lim_{x \to -3} \left( -2 x^{2} - 3 \right) = -21
The result
lim⁡x→−3(−2x2−3)=−21\lim_{x \to -3} \left( -2 x^{2} - 3 \right) = -21

Proved. Given any tolerance ε>0\varepsilon > 0, the choice δ=min⁡{1,  ε14}\delta = \min\left\{ 1, \; \frac{\varepsilon}{14} \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 = -21L − εc − δc = -3c + 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=−3c = -3, L=−21L = -21, ε=1100\varepsilon = \frac{1}{100} and δ=11400\delta = \frac{1}{1400}. 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=−3x = -3 into the function:

−2(−3)2−3\displaystyle -2 \left( -3 \right)^{2} - 3==−2⋅9−3\displaystyle -2 \cdot 9 - 3
Work out each power of −3-3 first. Powers come before multiplying, and multiplying before adding. checked
==−18−3\displaystyle -18 - 3
Do each multiplication. checked
==−21\displaystyle -21
Add the numbers up. checked
Claim: lim⁡x→−3(−2x2−3)=−21\text{Claim: } \lim_{x \to -3} \left( -2 x^{2} - 3 \right) = -21
This is what we will prove, with c=−3c = -3 and L=−21L = -21.
2

Rewrite everything in terms of the distance from c

Why this step

We control one thing only: how far xx is from −3-3. But ff is written in powers of xx, which hides that distance. So we give the distance a name, h=x−(−3)h = x - \left( -3 \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−(−3)h = x - \left( -3 \right) we get x=−3+hx = -3 + h. The condition 0<∣x−c∣<δ0 < \left| x - c \right| < \delta becomes simply 0<∣h∣<δ0 < \left| h \right| < \delta.

−2x2−3−(−21)\displaystyle -2 x^{2} - 3 - \left( -21 \right)==−2(−3+h)2−3−(−21)\displaystyle -2 \left( -3 + h \right)^{2} - 3 - \left( -21 \right)
Replace every xx by −3+h-3 + h. checked
==−2(9−6h+h2)−3−(−21)\displaystyle -2 \left( 9 - 6 h + h^{2} \right) - 3 - \left( -21 \right)
Multiply out the square: (u+h)2=u2+2uh+h2(u + h)^2 = u^2 + 2uh + h^2 with u=−3u = -3. checked
==−2h2+12h−18−3−(−21)\displaystyle -2 h^{2} + 12 h - 18 - 3 - \left( -21 \right)
Multiply each bracket by the number in front of it. A minus sign in front reaches every term inside. checked
==−2h2+12h\displaystyle -2 h^{2} + 12 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(−3)−Lf(-3) - L, and we CHOSE L=f(−3)L = f(-3). A wrong candidate for LL would leave a number here that no δ\delta could ever shrink. checked
==h(−2h+12)\displaystyle h \left( -2 h + 12 \right)
Every remaining term contains hh, so take hh out as a common factor. checked
∣−2x2−3−(−21)∣=∣h∣⋅∣−2h+12∣\left| -2 x^{2} - 3 - \left( -21 \right) \right| = \left| h \right| \cdot \left| -2 h + 12 \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