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→1lim(−x3−6x−2)=−9
The result
x→1lim(−x3−6x−2)=−9
Proved. Given any tolerance ε>0, the choice δ=min{1,13ε} 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=1, L=−9, ε=1001 and δ=13001. 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=1 into the function:
−13−6⋅1−2=−1−6⋅1−2
Work out each power of 1 first. Powers come before multiplying, and multiplying before adding. checked
=−1−6−2
Do each multiplication. checked
=−9
Add the numbers up. checked
Claim: x→1lim(−x3−6x−2)=−9
This is what we will prove, with c=1 and L=−9.
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=1, L=−9, ε=1001 and δ=13001. 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 1. But f is written in powers of x, which hides that distance. So we give the distance a name, h=x−1, and rewrite the function in powers of h. After this the quantity we control is the only letter on the page.
From h=x−1 we get x=1+h. The condition 0<∣x−c∣<δ becomes simply 0<∣h∣<δ.
−x3−6x−2−(−9)=−(1+h)3−6(1+h)−2−(−9)
Replace every x by 1+h. checked
=−(1+3h+3h2+h3)−6(1+h)−2−(−9)
Multiply out the powers: (u+h)2=u2+2uh+h2 and (u+h)3=u3+3u2h+3uh2+h3, with u=1. checked
=−h3−3h2−3h−1−6h−6−2−(−9)
Multiply each bracket by the number in front of it. A minus sign in front reaches every term inside. checked
=−h3−3h2−9h
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(1)−L, and we CHOSE L=f(1). A wrong candidate for L would leave a number here that no δ could ever shrink. checked
=h(−h2−3h−9)
Every remaining term contains h, so take h out as a common factor. checked
−x3−6x−2−(−9)=∣h∣⋅−h2−3h−9
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=1, L=−9, ε=101 and δ=1301. The bright part of the graph is the part inside the strip. The hollow circle marks the point itself, which the definition leaves out.