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(x3−x2+2x)=8
The result
x→2lim(x3−x2+2x)=8
Proved. Given any tolerance ε>0, the choice δ=min{1,16ε} 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=8, ε=1001 and δ=16001. 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:
23−22+2⋅2=8−4+2⋅2
Work out each power of 2 first. Powers come before multiplying, and multiplying before adding. checked
=8−4+4
Do each multiplication. checked
=8
Add the numbers up. checked
Claim: x→2lim(x3−x2+2x)=8
This is what we will prove, with c=2 and L=8.
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=8, ε=1001 and δ=16001. 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.
x3−x2+2x−8=(2+h)3−(2+h)2+2(2+h)−8
Replace every x by 2+h. checked
=(8+12h+6h2+h3)−(4+4h+h2)+2(2+h)−8
Multiply out the powers: (u+h)2=u2+2uh+h2 and (u+h)3=u3+3u2h+3uh2+h3, with u=2. checked
=h3+6h2+12h+8−h2−4h−4+2h+4−8
Multiply each bracket by the number in front of it. A minus sign in front reaches every term inside. checked
=h3+5h2+10h
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(h2+5h+10)
Every remaining term contains h, so take h out as a common factor. checked
x3−x2+2x−8=∣h∣⋅h2+5h+10
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=8, ε=101 and δ=1601. The bright part of the graph is the part inside the strip. The hollow circle marks the point itself, which the definition leaves out.