Why Prove the Sum and Difference Laws?
A limit law lets us find the limit of a sum or difference from the limits of its parts. The epsilon-delta proof explains why this works: if each function stays sufficiently close to its own limit, then their sum or difference stays close to the corresponding sum or difference of those limits.
The key is to coordinate two requirements. We choose a distance from the input point that works for both functions, and we divide the allowed output error between them. The triangle inequality then combines the two error bounds.
The Epsilon-Delta Definition
For a function \(f\) and real numbers \(a\) and \(L\), the statement \(\lim_{x\to a}f(x)=L\) means that every positive error tolerance can be met by taking \(x\) sufficiently close to \(a\), without requiring \(x=a\).
The definition concerns values of \(x\) near \(a\), not necessarily the value of \(f\) at \(a\). In the proofs below, assume that the two functions are defined on a common punctured neighborhood of \(a\), so both can be evaluated whenever \(x\) is sufficiently close to \(a\) and \(x\ne a\).
The Sum and Difference Laws
Suppose that \(\lim_{x\to a}f(x)=L\) and \(\lim_{x\to a}g(x)=M\), where \(L\) and \(M\) are finite real numbers. The sum law says that the limit of \(f+g\) is \(L+M\); the difference law says that the limit of \(f-g\) is \(L-M\). We now prove both directly from the definition.
Why Splitting the Tolerance Works
To prove either law, fix an arbitrary \(\varepsilon>0\), the error tolerance for the combined expression. For the sum, the triangle inequality gives
Thus it is enough to make each error on the right less than \(\varepsilon/2\). Their sum will then be less than \(\varepsilon\). The difference proof has the same error bound, because
Each limit hypothesis supplies a suitable distance for its own function at tolerance \(\varepsilon/2\). Those distances need not be equal. Taking the smaller one ensures that both bounds hold at once.
Let \(\varepsilon>0\) be arbitrary.
Use tolerance \(\varepsilon/2\) for the error in \(f\) and \(\varepsilon/2\) for the error in \(g\).
Take the minimum of the two positive distances supplied by the limit hypotheses.
Apply the triangle inequality; the two errors add to less than \(\varepsilon\).
Complete Proof of Both Laws
Fix \(\varepsilon>0\). Since \(\lim_{x\to a}f(x)=L\), the definition applied with tolerance \(\varepsilon/2\) gives a number \(\delta_f>0\) such that
Likewise, since \(\lim_{x\to a}g(x)=M\), there is a number \(\delta_g>0\) such that
Set \(\delta=\min(\delta_f,\delta_g)\). This is positive. If \(0<|x-a|<\delta\), then \(0<|x-a|<\delta_f\) and \(0<|x-a|<\delta_g\). Consequently, both error bounds hold. For the sum,
This verifies the epsilon-delta definition for the sum, so \(\lim_{x\to a}(f(x)+g(x))=L+M\).
For the difference, with the same \(\delta\),
This verifies the definition for the difference, so \(\lim_{x\to a}(f(x)-g(x))=L-M\). Since \(\varepsilon>0\) was arbitrary, both conclusions follow.
Worked Example: Proving a Sum of Linear Functions
Worked Example: Proving a Sum of Linear Functions
Use the definition and tolerance splitting to prove that \(\lim_{x\to 2}\big((2x-1)+(x+4)\big)=9\).
At \(x=2\), the two component values are \(2(2)-1=3\) and \(2+4=6\), whose sum is \(9\). Let \(\varepsilon>0\). For the first function,
To make this less than \(\varepsilon/2\), it is enough to require \(|x-2|<\varepsilon/4\). For the second function,
which is less than \(\varepsilon/2\) whenever \(|x-2|<\varepsilon/2\). Choose \(\delta=\varepsilon/4\). Then \(0<|x-2|<\delta\) implies both bounds, since \(\varepsilon/4\leq\varepsilon/2\). Therefore,
Check the target by simplifying the expression: \((2x-1)+(x+4)=3x+3\), and \(3(2)+3=9\). The estimate above proves that its values approach \(9\) in the required epsilon-delta sense.
Worked Example: Proving a Difference with a Quadratic Term
Worked Example: Proving a Difference with a Quadratic Term
Prove that \(\lim_{x\to 1}(x^2-2x)=-1\).
The component limits at \(x=1\) are \(1^2=1\) and \(2(1)=2\), so their difference is \(1-2=-1\). Let \(\varepsilon>0\). For the first component, factor the error:
Require \(|x-1|<1\). Then \(0<x<2\), so \(1<x+1<3\), and hence \(|x+1|<3\). It follows that \(|x^2-1|<3|x-1|\). Thus \(|x^2-1|<\varepsilon/2\) if \(|x-1|<\varepsilon/6\). For the second component,
so \(|2x-2|<\varepsilon/2\) if \(|x-1|<\varepsilon/4\). Choose \(\delta=\min(1,\varepsilon/6)\). This choice enforces \(|x-1|<1\) and \(|x-1|<\varepsilon/6\); also \(\delta\leq\varepsilon/6\leq\varepsilon/4\), so it enforces the second component’s bound as well.
For \(0<|x-1|<\delta\), the triangle inequality now gives
As a check, the difference from the proposed limit simplifies to \((x^2-2x)+1=(x-1)^2\), which equals \(0\) at \(x=1\). The inequalities establish that this difference is less than any prescribed positive tolerance when \(x\) is sufficiently close to \(1\).
Worked Example: Proving a Sum with Two Different Bounds
Worked Example: Proving a Sum with Two Different Bounds
Prove that \(\lim_{x\to 3}\big(x^2+(5-x)\big)=11\).
The component values at \(x=3\) are \(9\) and \(2\), giving \(9+2=11\). Let \(\varepsilon>0\). For the quadratic component,
If \(|x-3|<1\), then \(2<x<4\), and therefore \(5<x+3<7\), so \(|x+3|<7\). Thus \(|x^2-9|<7|x-3|\), which is less than \(\varepsilon/2\) if \(|x-3|<\varepsilon/14\). For the other component,
which is less than \(\varepsilon/2\) whenever \(|x-3|<\varepsilon/2\). Choose \(\delta=\min(1,\varepsilon/14)\). It ensures the quadratic estimate, and since \(\delta\leq\varepsilon/14\leq\varepsilon/2\), it also ensures the second estimate. Therefore,
Finally, \(x^2+(5-x)=x^2-x+5\), and substituting \(x=3\) gives \(9-3+5=11\), confirming the stated limit value.
Common Mistake: Spending the Full Tolerance Twice
A tempting but insufficient approach is to use the full tolerance \(\varepsilon\) for each component. That only proves that the combined error is less than \(2\varepsilon\), not necessarily less than \(\varepsilon\).
For a concrete failure, take \(f(x)=g(x)=x\), with \(x\to0\), and set \(\varepsilon=0.1\). If one uses \(\delta=0.1\), then \(x=0.075\) satisfies \(0<|x|<\delta\). Each component error is \(0.075<0.1\), but the sum error is
The correction is to assign half the tolerance to each component. For this example, choosing \(\delta=0.05\) ensures that \(0<|x|<\delta\) gives \(|f(x)|<0.05\) and \(|g(x)|<0.05\). Hence \(|f(x)+g(x)|<0.05+0.05=0.1\), as required.
What to Keep in Mind
The proof rests on two precise choices. First, split the desired error \(\varepsilon\) into positive allowances whose sum is \(\varepsilon\); using \(\varepsilon/2\) for each function is the simplest choice. Second, take the minimum of the distances provided by the two limit hypotheses, so that a single condition on \(x\) makes both component estimates valid.
The same argument handles a difference because subtracting the second component changes the expression but not the error bound: the triangle inequality still bounds the total by the sum of the two component errors. The argument requires both component limits to be finite and to be defined for the same nearby input values. It does not require either function to be defined at \(a\).
Check Your Understanding
Answer these questions by referring to the definition and estimates in the proof.
- Why is \(\delta=\min(\delta_f,\delta_g)\) used rather than choosing only one of the two distances?
- In the sum proof, what bound follows when each component error is less than \(\varepsilon/2\)?
- Why does the difference proof also use the sum of the two component errors?
- For the functions \(f(x)=g(x)=x\) near \(0\), what delta choice makes each error less than \(\varepsilon/2\)?
- Does the proof require \(f(a)\) and \(g(a)\) to be defined? Explain briefly.
