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
Prove this limit using only the epsilon-delta definition, then try the proof out with the tolerance ε=1001.
x→2lim(2x3−2x2−3x+4)=6
The result
x→2lim(2x3−2x2−3x+4)=6
Proved. Given any tolerance ε>0, the choice δ=min{1,25ε} works.
checked18 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
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=2, L=6, ε=1001 and δ=25001. 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=2 into the function:
2⋅23−2⋅22−3⋅2+4=2⋅8−2⋅4−3⋅2+4
Work out each power of 2 first. Powers come before multiplying, and multiplying before adding. checked
=16−8−6+4
Do each multiplication. checked
=6
Add the numbers up. checked
Claim: x→2lim(2x3−2x2−3x+4)=6
This is what we will prove, with c=2 and L=6.
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=2, L=6, ε=1001 and δ=25001. 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
Rewrite everything in terms of the distance from c
Why this step
We control one thing only: how far x is from 2. But f is written in powers of x, which hides that distance. So we give the distance a name, h=x−2, and rewrite the function in powers of h. After this the quantity we control is the only letter on the page.
From h=x−2 we get x=2+h. The condition 0<∣x−c∣<δ becomes simply 0<∣h∣<δ.
2x3−2x2−3x+4−6=2(2+h)3−2(2+h)2−3(2+h)+4−6
Replace every x by 2+h. checked
=2(8+12h+6h2+h3)−2(4+4h+h2)−3(2+h)+4−6
Multiply out the powers: (u+h)2=u2+2uh+h2 and (u+h)3=u3+3u2h+3uh2+h3, with u=2. checked
=2h3+12h2+24h+16−2h2−8h−8−3h−6+4−6
Multiply each bracket by the number in front of it. A minus sign in front reaches every term inside. checked
=2h3+10h2+13h
Collect the terms with the same power of h. The plain numbers add up to exactly 0, and that is no accident: they are f(2)−L, and we CHOSE L=f(2). A wrong candidate for L would leave a number here that no δ could ever shrink. checked
=h(2h2+10h+13)
Every remaining term contains h, so take h out as a common factor. checked
2x3−2x2−3x+4−6=∣h∣⋅2h2+10h+13
Take absolute values. The absolute value of a product is the product of the absolute values. checked
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=2, L=6, ε=101 and δ=2501. The bright part of the graph is the part inside the strip. The hollow circle marks the point itself, which the definition leaves out.