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(−4x2−7x−5)=−35
The result
x→2lim(−4x2−7x−5)=−35
Proved. Given any tolerance ε>0, the choice δ=min{1,27ε} 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=−35, ε=1001 and δ=27001. 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:
−4⋅22−7⋅2−5=−4⋅4−7⋅2−5
Work out each power of 2 first. Powers come before multiplying, and multiplying before adding. checked
=−16−14−5
Do each multiplication. checked
=−35
Add the numbers up. checked
Claim: x→2lim(−4x2−7x−5)=−35
This is what we will prove, with c=2 and L=−35.
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=−35, ε=1001 and δ=27001. 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.
−4x2−7x−5−(−35)=−4(2+h)2−7(2+h)−5−(−35)
Replace every x by 2+h. checked
=−4(4+4h+h2)−7(2+h)−5−(−35)
Multiply out the square: (u+h)2=u2+2uh+h2 with u=2. checked
=−4h2−16h−16−7h−14−5−(−35)
Multiply each bracket by the number in front of it. A minus sign in front reaches every term inside. checked
=−4h2−23h
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(−4h−23)
Every remaining term contains h, so take h out as a common factor. checked
−4x2−7x−5−(−35)=∣h∣⋅∣−4h−23∣
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=−35, ε=101 and δ=2701. The bright part of the graph is the part inside the strip. The hollow circle marks the point itself, which the definition leaves out.