From a Tolerance to a Delta
A quadratic limit proof can be made concrete by starting with a particular error tolerance and producing a specific input distance that guarantees it. The main work is to factor the difference between the function value and its proposed limit, then bound the factor that remains.
For a quadratic function, that remaining factor usually depends on the input \(x\). A bound that is valid only near the point of interest is enough, provided the chosen delta keeps \(x\) in that neighbourhood. The proof must make both parts explicit.
Here, \(\varepsilon\) is the allowed error in the function value, and \(\delta\) is the input distance that guarantees that error. In this tutorial, the tolerance is a specified positive number, so the proof needs to provide and verify a numerical choice of \(\delta\).
A Local Bound for a Quadratic
Let \(q(x)=Ax^2+Bx+C\), where \(A\ne 0\), and consider the limit at \(x=a\). The proposed limit value is \(q(a)\). Subtracting the two values and factoring gives:
This factorization separates the input distance \(|x-a|\) from a second factor, \(|A(x+a)+B|\), which varies with \(x\). To control the second factor, restrict \(x\) to within distance \(1\) of \(a\). If \(|x-a|<1\), then the triangle inequality gives:
Applying the triangle inequality once more produces the local bound:
Write \(M=|A|(2|a|+1)+|B|\). Since \(A\ne 0\), \(M>0\). The displayed inequalities show that whenever \(|x-a|<1\), the changing factor is less than \(M\). Thus the quadratic difference is bounded by \(M|x-a|\).
For a specified \(\varepsilon>0\), a general choice is \(\delta=\min(1,\varepsilon/M)\). The first restriction keeps \(x\) within the neighbourhood where the bound applies; the second makes the error small enough. The examples below carry out this reasoning with particular tolerances.
Worked Proofs with Numerical Tolerances
Worked Example: A Tolerance of One Tenth for \(x^2\) at \(2\)
Prove that if \(0<|x-2|<\delta\), then \(|x^2-4|<0.1\), using an explicit value of \(\delta\).
First factor the difference from the proposed limit value \(4\):
Use the neighbourhood \(|x-2|<1\). The triangle inequality gives \(|x+2|\le |x-2|+4<5\). Therefore, within this neighbourhood, \(|x^2-4|<5|x-2|\). To make this less than \(0.1\), choose \(\delta=0.1/5=0.02\), which is also less than \(1\).
Now verify the choice. If \(0<|x-2|<0.02\), then \(|x-2|<1\), so \(|x+2|<5\). Hence:
Thus \(\delta=0.02\) guarantees the required error tolerance. The factorization also checks that the target value is correct: substituting \(x=2\) gives \(2^2=4\), and the difference factors as \((x-2)(x+2)\).
Worked Example: A Tolerance of \(0.45\) for \(2x^2-3x+1\) at \(-1\)
Prove that \(\lim_{x\to -1}(2x^2-3x+1)=6\) for the specified tolerance \(\varepsilon=0.45\).
Evaluate the function at the input \(-1\):
Subtract this value and factor. The difference-of-squares identity gives:
Restrict first to \(|x+1|<1\). Then \(|x|=|x+1-1|\le |x+1|+1<2\), so \(|2x-5|\le 2|x|+5<9\). Consequently, the function-value error is less than \(9|x+1|\). Set \(\delta=0.45/9=0.05\), which is less than \(1\).
To verify this delta, suppose \(0<|x+1|<0.05\). This implies \(|x+1|<1\), so the bound \(|2x-5|<9\) applies. Therefore:
Every inequality follows from the assumed input distance and the local bound. The proposed limit value is checked by the direct evaluation at \(-1\), and the chosen \(\delta=0.05\) proves the required claim.
Worked Example: A Tolerance of \(0.3\) for \(-x^2+4x-2\) at \(3\)
Prove that \(\lim_{x\to 3}(-x^2+4x-2)=1\) for \(\varepsilon=0.3\). This example uses a direct local bound for the factor after factoring.
Check the proposed limit value:
The difference from this value factors as follows:
When \(|x-3|<1\), we have \(2<x<4\). Therefore \(-3<1-x<-1\), which implies \(|1-x|<3\). It follows that the error is less than \(3|x-3|\). Choose \(\delta=0.3/3=0.1\), which is less than \(1\).
For the verification, suppose \(0<|x-3|<0.1\). Then \(|x-3|<1\), so \(2<x<4\) and \(|1-x|<3\). Thus:
So \(\delta=0.1\) works. The proof checks both requirements: the input stays in the neighbourhood where the factor is bounded, and the resulting error is strictly below the specified tolerance.
Why the Bound and the Neighbourhood Belong Together
The quadratic difference contains a factor that depends on \(x\). A numerical bound for that factor is justified only on a stated range of inputs. In the examples, the neighbourhood restriction makes the factor bounded, and the delta is chosen to satisfy both the neighbourhood restriction and the error requirement.
This is the reason a minimum is often used in the general choice \(\delta=\min(1,\varepsilon/M)\). The value \(1\) keeps the input inside the region used to establish the factor bound. The value \(\varepsilon/M\) makes the product of the bound and the input distance smaller than \(\varepsilon\). If the smaller value is used as \(\delta\), both conditions hold.
A Common Mistake: Using a Bound Without Its Conditions
Consider the first example again. It would be incorrect to claim that \(|x+2|<4\) for all \(x\). For instance, at \(x=3\), \(|x+2|=5\). A bound for all real inputs cannot be inferred just because the factor has a particular value at the limit input.
That mistaken bound might lead to the proposed choice \(\delta=0.1/4=0.025\). It does not guarantee the tolerance. Take \(x=2.0249\). Then:
So the proposed delta fails for an input that satisfies its condition. The correction is to justify the bound on a neighbourhood: if \(|x-2|<1\), then \(|x+2|<5\). The resulting choice \(\delta=0.02\) keeps \(x\) in that neighbourhood and gives the verified error bound \(5(0.02)=0.1\), with a strict inequality because \(|x-2|<\delta\).
Why Concrete Tolerances Matter
Working with a specific \(\varepsilon\) makes the role of each part of the proof visible. The tolerance sets the required size of the final error; the factorization reveals how the input distance influences that error; and a local bound turns the changing factor into a fixed number. The delta is not guessed from a graph or chosen merely because it looks small: it is supported by the inequalities in the proof.
The same reasoning applies to different quadratic functions and different input points, though the local factor bound changes. A verified delta proves the required implication for every input satisfying \(0<|x-a|<\delta\), not just for a few test inputs.
Check Your Understanding
For each question, identify the factorization or bound needed before choosing a delta.
- For \(x^2\) at \(1\), factor \(|x^2-1|\) and find a bound for the second factor when \(|x-1|<1\).
- For \(q(x)=x^2+2x\) at \(0\), compute \(q(0)\) and factor \(|q(x)-q(0)|\).
- In the general quadratic estimate, what two conditions does \(\delta=\min(1,\varepsilon/M)\) enforce?
- Why must a bound on the factor depending on \(x\) be justified on a neighbourhood of the input?
- What is the difference between checking the error at a single input and verifying the epsilon-delta implication?
