Why Sum and Difference Laws Matter
Often, the limit of a complicated expression is easier to find by separating it into simpler pieces. If the limits of those pieces are already known, the sum and difference laws let us combine the results without evaluating the original expression by direct substitution.
The key requirement is that the limits of the separate functions must exist as finite real numbers at the same approach point. The functions do not need to have values at that point, and their values there need not equal their limits.
The Sum and Difference Laws
The requirement that both limits be finite matters. These laws, as stated here, do not cover expressions involving infinite limits. The requirement that the approach point be the same matters too: knowing limits of \(f\) and \(g\) at different points does not determine the limit of \(f+g\) at either point.
It is also important to distinguish the limit from the function's value at \(a\). A limit describes the function's behavior as \(x\) approaches \(a\), not necessarily what happens at \(x=a\). The laws require the functions to be defined near \(a\), but they do not require either function to be defined at \(a\).
Why the Laws Work: An Error Estimate
Suppose \(f(x)\) is close to \(L\), and \(g(x)\) is close to \(M\). The errors in the two approximations are \(f(x)-L\) and \(g(x)-M\). The error in approximating their sum by \(L+M\) is exactly
The triangle inequality therefore gives
Thus, if both component errors are small, the sum's error is small as well. For a difference, the corresponding identity is
and the triangle inequality again bounds the absolute error by \(|f(x)-L|+|g(x)-M|\). This error estimate explains why the laws hold: sufficiently close component values produce a sufficiently close sum or difference. A formal proof translates this control into the precise definition of a limit.
Worked Examples
Worked Example: Adding Limits When Point Values Mislead
Define \(f\) and \(g\) near \(x=2\) by
Find \(\lim_{x\to 2}(f(x)+g(x))\), and compare it with the value of \(f(2)+g(2)\).
For \(x\ne 2\), \(f(x)=x+4\). As \(x\) approaches 2, \(x+4\) approaches \(2+4=6\), so \(\lim_{x\to 2}f(x)=6\). Also, for \(x\ne 2\), \(g(x)=2-x\), which approaches \(2-2=0\). Therefore, \(\lim_{x\to 2}g(x)=0\).
Both component limits are finite and taken at the same point, so the sum law applies:
To check this result directly near, but not at, 2, use the definitions for \(x\ne 2\):
Thus the sum is exactly 6 for every \(x\ne 2\) in this neighborhood, and its limit is 6. At the point itself, however, \(f(2)+g(2)=100+(-50)=50\). The function value is 50, while the limit is 6. This confirms why substituting the point values would give the wrong limit.
Worked Example: Subtracting Limits with a Different Point Value
Let
Find \(\lim_{x\to 1}(p(x)-q(x))\).
For \(x\ne 1\), \(p(x)=x^2\), so \(\lim_{x\to 1}p(x)=1^2=1\). Also, \(q(x)=3x-2\) when \(x\ne 1\), so \(\lim_{x\to 1}q(x)=3(1)-2=1\).
Since both finite limits exist at 1, the difference law gives
For a direct check, if \(x\ne 1\), then
The difference between this expression and the proposed limit 0 factors as
When \(|x-1|<1\), we have \(0<x<2\), so \(|x-2|<2\). Hence, for \(x\ne 1\) with \(|x-1|<1\),
As \(x\) approaches 1, this bound approaches 0, confirming the limit. At \(x=1\), the difference is \(p(1)-q(1)=7-10=-3\), not 0. Again, the point value does not determine the limit.
Worked Example: Combining Three Known Limits
Suppose that, as \(x\to -1\), the finite limits of \(u(x)\), \(v(x)\), and \(w(x)\) are 3, 2, and 5, respectively. Find
First apply the sum law to \(u+v\): its limit is \(3+2=5\). Both that limit and the limit of \(w\) exist and are finite at \(-1\), so apply the difference law:
For a numerical check of the arithmetic, the result of the first operation is \(3+2=5\), and the second is \(5-5=0\). The order of the operations matches the grouping in the expression. This is repeated use of the two laws, not a separate rule: each application requires the finite limits of the two functions being combined.
Applying the Laws Carefully
For an expression with several additions and subtractions, apply the laws in stages. At each stage, check that the limits being combined are finite and refer to the same approach point. For example, once the limits of \(u\), \(v\), and \(w\) at \(a\) are known, the limit of \((u+v)-w\) is found by first combining the limits of \(u\) and \(v\), then subtracting the limit of \(w\).
The laws are implications, not guarantees that a limit exists in every situation. They say that if the separate finite limits exist, then the limit of the sum or difference exists and has the stated value. They do not say that the component limits must exist whenever the sum or difference has a limit.
For example, for \(x\ne 0\), define \(r(x)=1/x\) and \(s(x)=-1/x\). Neither function has a finite limit as \(x\to 0\). Yet \(r(x)+s(x)=0\) for every \(x\ne 0\), so the sum has limit 0. The sum law cannot be used to establish this result because its required component limits are absent; here, the cancellation is seen directly in the combined expression.
What to Remember
The sum and difference laws turn known finite limits into new ones: add the limits when adding the functions, and subtract the limits when subtracting the functions. Their assumptions are precise: both component limits must exist as finite real numbers and must be taken at the same approach point. Function values at that point are irrelevant to the laws.
Check Your Understanding
Use the conditions of the laws as well as the arithmetic when answering these questions.
- If \(\lim_{x\to 3}f(x)=4\) and \(\lim_{x\to 3}g(x)=-2\), what is \(\lim_{x\to 3}(f(x)+g(x))\)?
- Under the same assumptions, what is \(\lim_{x\to 3}(f(x)-g(x))\)?
- Must \(f(a)\) equal \(\lim_{x\to a}f(x)\) for the sum law to apply? Explain.
- What condition on the component limits is needed before applying the difference law?
- Can the sum have a finite limit even if one or both component limits do not exist? Give a reason for your answer.
