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(−2x2+5x+1)=−6
The result
x→−1lim(−2x2+5x+1)=−6
Proved. Given any tolerance ε>0, the choice δ=min{1,11ε} 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=−6, ε=1001 and δ=11001. 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:
−2(−1)2+5(−1)+1=−2⋅1+5(−1)+1
Work out each power of −1 first. Powers come before multiplying, and multiplying before adding. checked
=−2−5+1
Do each multiplication. checked
=−6
Add the numbers up. checked
Claim: x→−1lim(−2x2+5x+1)=−6
This is what we will prove, with c=−1 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=−1, L=−6, ε=1001 and δ=11001. 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.
−2x2+5x+1−(−6)=−2(−1+h)2+5(−1+h)+1−(−6)
Replace every x by −1+h. checked
=−2(1−2h+h2)+5(−1+h)+1−(−6)
Multiply out the square: (u+h)2=u2+2uh+h2 with u=−1. checked
=−2h2+4h−2+5h−5+1−(−6)
Multiply each bracket by the number in front of it. A minus sign in front reaches every term inside. checked
=−2h2+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(−2h+9)
Every remaining term contains h, so take h out as a common factor. checked
−2x2+5x+1−(−6)=∣h∣⋅∣−2h+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=−6, ε=101 and δ=1101. The bright part of the graph is the part inside the strip. The hollow circle marks the point itself, which the definition leaves out.