From a Quadratic Difference to a Proof
A quadratic limit proof is built around a simple observation: the difference between a quadratic value and its value at a specified input contains the distance from that input as a factor. The remaining factor may vary with \(x\), so a complete proof must also control its size near the input.
For example, if \(q(x)=3x^2-2x+1\) and the input is \(r=2\), then \(q(2)=9\), and factoring gives \(q(x)-q(2)=(x-2)(3x+4)\). The factor \(x-2\) measures how close \(x\) is to 2. The proof will make the other factor bounded whenever \(x\) is sufficiently close to 2.
For a quadratic limit at \(r\), the proposed value will be \(q(r)\). The task is to show that an arbitrary positive error tolerance \(\varepsilon\) can be met by choosing a positive distance \(\delta\). A proof must show why that choice works, not merely state it.
A General Bound for a Quadratic
Let \(q(x)=Ax^2+Bx+C\), where \(A,B,C\) are fixed real numbers, and let \(r\) be any real input. Subtracting \(q(r)\) and factoring gives
The constant term \(C\) disappears in the subtraction. To control the error, we need a bound on \(|A(x+r)+B|\). Restrict first to \(|x-r|<1\). Since \(x+r=(x-r)+2r\), the triangle inequality gives
Therefore, another use of the triangle inequality gives
Define the nonnegative number \(M=|A|(1+2|r|)+|B|\). On the restriction \(|x-r|<1\), the factor is at most \(M\), and hence also at most \(M+1\). Using \(M+1\) keeps the eventual choice valid even when \(M=0\).
Why the Choice of Delta Works
The proof below checks both restrictions built into the choice of \(\delta\): it is no greater than 1, which supports the local bound, and no greater than \(\varepsilon/(M+1)\), which makes the final error small.
Let \(\varepsilon>0\). Define \(M=|A|(1+2|r|)+|B|\), and choose \(\delta=\min\{1,\varepsilon/(M+1)\}\). Both entries in the minimum are positive, so \(\delta>0\).
Suppose \(0<|x-r|<\delta\). Since \(\delta\le 1\), we have \(|x-r|<1\), so the local estimate gives \(|A(x+r)+B|\le M+1\).
Factor the difference and apply the two bounds: \(|q(x)-q(r)|=|x-r||A(x+r)+B|\le |x-r|(M+1)\).
Because \(|x-r|<\delta\le\varepsilon/(M+1)\), it follows that \(|q(x)-q(r)|<\varepsilon\). This proves the required implication for every \(\varepsilon>0\).
This is a direct epsilon-delta proof: it begins with an arbitrary positive tolerance, supplies a positive \(\delta\), and verifies that every \(x\) satisfying the required distance condition produces an error below that tolerance. The estimate works for every real \(r\); no special assumption about whether \(r\) is positive, negative, or zero is needed.
Worked Proofs with Concrete Tolerances
Worked Example: A Quadratic at a Positive Input
Prove from the definition that \(\lim_{x\to 2}(3x^2-2x+1)=9\) for \(\varepsilon=1/10\).
First evaluate the proposed limit value: \(q(2)=3(2^2)-2(2)+1=3(4)-4+1=12-4+1=9\). Factor the error by subtracting this value:
To bound the second factor, require \(|x-2|<1\). Then \(-1<x-2<1\), so \(1<x<3\). Thus \(7<3x+4<13\), which implies \(|3x+4|<13\). Choose \(\delta=\min\{1,(1/10)/13\}=1/130\). If \(0<|x-2|<\delta\), then
The factorization is correct because \((x-2)(3x+4)=3x^2-2x-8\), and adding 9 recovers \(q(x)\). The chosen \(\delta\) therefore proves the required claim for this tolerance.
Worked Example: A Quadratic at a Different Input
Prove from the definition that \(\lim_{x\to 1}(2x^2+x-3)=0\) for \(\varepsilon=1/5\).
Evaluate \(q(1)=2(1^2)+1-3=2+1-3=0\). Subtracting that value and factoring gives
For the last equality, \((x-1)(2x+3)=2x^2+x-3\). If \(|x-1|<1\), then \(0<x<2\), so \(3<2x+3<7\), and consequently \(|2x+3|<7\). Set \(\delta=\min\{1,(1/5)/7\}=1/35\). Whenever \(0<|x-1|<1/35\), the local condition \(|x-1|<1\) holds and
Substituting \(x=1\) into the original function gives the proposed value 0, and the displayed estimate verifies the tolerance. Thus the proof is complete.
Worked Example: A Quadratic at a Negative Input
Prove from the definition that \(\lim_{x\to -1}(-2x^2+5x+1)=-6\) for \(\varepsilon=1/2\).
Evaluate \(q(-1)=-2((-1)^2)+5(-1)+1=-2(1)-5+1=-6\). Subtract \(-6\) and factor:
Indeed, \((x+1)(7-2x)=7x+7-2x^2-2x=-2x^2+5x+7\). If \(|x+1|<1\), then \(-2<x<0\). It follows that \(7<7-2x<11\), so \(|7-2x|<11\). Choose \(\delta=\min\{1,(1/2)/11\}=1/22\). For \(0<|x+1|<1/22\), both the local restriction and the following estimate hold:
The substitution check gives \(q(-1)=-6\), and the inequality establishes the desired error bound. The argument also confirms why the sign of the input causes no difficulty: the local interval here is explicitly \((-2,0)\).
Common Mistakes and Their Corrections
A frequent error is to treat the factor that depends on \(x\) as though it were automatically constant. For example, in the first example, the following calculation is not justified:
The bound \(|3x+4|\le 7\) does not hold for all real \(x\). For instance, at \(x=3\), \(|3x+4|=13\), which is greater than 7. A bound must be connected to a stated restriction on \(x\), and it must be true under that restriction.
The correction is to require \(|x-2|<1\). This implies \(1<x<3\), and therefore \(7<3x+4<13\). A valid bound is then \(|3x+4|<13\). Choosing \(\delta\le1\) ensures that this local restriction is in force whenever \(|x-2|<\delta\).
Another mistake is to choose \(\delta=\varepsilon\) just because the difference has a factor \(x-r\). That ignores the size of the other factor. If that factor can be larger than 1, the product can exceed \(\varepsilon\) even when \(|x-r|<\varepsilon\). First establish a local bound \(K\) for the remaining factor; then require \(\delta\le\varepsilon/K\) when \(K>0\), as well as any restriction needed to ensure the bound applies.
Conditions and a Reliable Checklist
The general result applies to every real quadratic \(q(x)=Ax^2+Bx+C\) and every real input \(r\). The coefficients are fixed real numbers, the tolerance must be any \(\varepsilon>0\), and the chosen \(\delta\) must be positive. The proof does not require \(A\ne0\): if \(A=0\), the expression is linear or constant, and the same factorization and bound remain valid. It also does not require \(x=r\), because the definition tests inputs with \(0<|x-r|<\delta\).
When constructing a proof for a particular quadratic, keep the reasoning in this order:
- Compute \(q(r)\) to identify the proposed limit.
- Subtract \(q(r)\) and factor the difference to expose \(x-r\).
- Choose a simple local restriction, commonly \(|x-r|<1\).
- Translate that restriction into an interval for \(x\) and bound the remaining factor.
- Choose a positive \(\delta\) that enforces both the local restriction and the required error tolerance.
- Assume \(0<|x-r|<\delta\) and write out the inequalities that lead to \(|q(x)-q(r)|<\varepsilon\).
Each step has a distinct role. Factoring identifies the small quantity; the local restriction controls the other factor; and the final inequality connects the chosen distance to the requested error. Leaving out any one of these links can make an otherwise plausible argument incomplete.
Check Your Understanding
For each question, make sure your reasoning includes the factorization and the inequalities needed to justify any local bound.
- For \(q(x)=x^2+2x\) at \(r=1\), calculate \(q(1)\) and factor \(q(x)-q(1)\).
- For \(q(x)=x^2-5x+4\) at \(r=2\), use \(|x-2|<1\) to find an upper bound for the absolute value of the factor remaining after \(x-2\) is removed.
- For \(q(x)=2x^2-3x+1\) at \(r=0\), write a valid positive \(\delta\) in terms of an arbitrary \(\varepsilon>0\), using the general theorem.
- Why does a bound on the remaining factor need a stated restriction on \(x\)? Give one specific reason based on a factor that grows as \(x\) changes.
- In a proof, what two properties of \(\delta\) are needed when the local estimate is established under \(|x-r|<1\)?
