Why Limit Laws Work
When two functions approach fixed values near the same input, their sum and product also approach predictable values. The sum approaches the sum of the individual limits, and the product approaches their product. These rules are useful because they let us evaluate a more complicated expression by understanding its simpler parts.
It is important that these are statements about nearby function values, not necessarily about the functions' values at the target input. A function can be undefined at that input, or its value there can differ from the value it approaches. What matters for a limit is the behavior as the input gets close to the target.
For the laws in this lesson, suppose \(f\) and \(g\) are defined for inputs sufficiently close to \(a\), apart from possibly at \(a\), and suppose both limits exist as finite real numbers. Then:
The sum law says that the limit of \(f(x)+g(x)\) is \(L+M\). The product law says that the limit of \(f(x)g(x)\) is \(LM\). We will justify both laws from the meaning of a limit, then use them in calculations.
The Sum Limit Law
The reason is that the difference between the sum and its proposed limit is the sum of the two individual differences. If each individual difference is small, their sum is small as well.
Let \(\varepsilon>0\) be the maximum allowed distance between \(f(x)+g(x)\) and \(L+M\).
Because \(f(x)\to L\) and \(g(x)\to M\), choose \(x\) close enough to \(a\) that \(|f(x)-L|<\varepsilon/2\) and \(|g(x)-M|<\varepsilon/2\).
The triangle inequality shows that \(|(f(x)+g(x))-(L+M)|<\varepsilon\), proving the stated limit.
Here is the full estimate behind the last step:
Each of the two limits supplies a neighborhood of \(a\) in which its required inequality holds. Taking \(x\) close enough to satisfy both inequalities at once completes the argument. Thus the sum law follows directly from the definition of limit and the triangle inequality.
The Product Limit Law
For a product, it is not enough simply to add the differences from the limiting values. First, we need to know that \(f(x)\) stays bounded near \(a\). The fact that \(f(x)\to L\) gives that bound: close enough to \(a\), we can require \(|f(x)-L|<1\), which implies \(|f(x)|<|L|+1\).
Now split the product's error into two parts. This identity is the key to the proof:
Taking absolute values and using the triangle inequality gives:
Let \(\varepsilon>0\). Choose \(x\) sufficiently close to \(a\) that all three inequalities hold:
Such a choice is possible because both functions approach their respective limits; we can take \(x\) close enough to satisfy all the conditions simultaneously. The first inequality gives \(|f(x)|<|L|+1\). Therefore:
This proves that \(f(x)g(x)\) approaches \(LM\), as claimed. The boundedness step matters: it ensures that the error in \(g(x)\) cannot be magnified without control by \(f(x)\).
Worked Examples
Worked Example: Add Two Known Limits
Suppose \(\lim_{x\to 2}f(x)=7\) and \(\lim_{x\to 2}g(x)=-4\). Find \(\lim_{x\to 2}\bigl(f(x)+g(x)\bigr)\).
Both limits exist and are finite, so the sum law applies. Add their limiting values:
Check: Near \(x=2\), \(f(x)\) is close to \(7\) and \(g(x)\) is close to \(-4\). Their sum is therefore close to \(7+(-4)=3\), in agreement with the result.
Worked Example: Multiply Two Known Limits
Suppose \(\lim_{x\to -1}u(x)=3\) and \(\lim_{x\to -1}v(x)=2\). Find \(\lim_{x\to -1}u(x)v(x)\).
The component limits are both finite, so the product law applies. Multiply the limiting values:
Check: The proposed product limit is the product of the two component limits. Substituting those limiting values into that product gives \(3\cdot 2=6\), as required by the product law.
Worked Example: Combine Sum and Product Laws
Suppose \(\lim_{x\to 4}r(x)=2\) and \(\lim_{x\to 4}s(x)=-3\). Find \(\lim_{x\to 4}\bigl(r(x)s(x)+r(x)\bigr)\).
First use the product law on \(r(x)s(x)\). Then use the sum law to add \(r(x)\):
Check: The product part approaches \(-6\), and the remaining term approaches \(2\). Their sum is \(-6+2=-4\), matching the result.
What the Laws Do—and Do Not—Say
The sum and product laws allow us to replace component functions by their limits when the conditions hold: each component limit must exist and be finite. The operations on the right side are ordinary addition and multiplication of real numbers.
These rules do not claim that the function values at \(x=a\) equal their limits. For example, the laws concern what happens as \(x\) approaches \(a\), and the functions may even be undefined at \(a\). Nor do the laws establish that a limit exists when the component limits are unknown. Their stated assumptions are essential.
A useful habit is to identify the outer operation before calculating. If the expression is a sum, apply the sum law; if it is a product, apply the product law. For an expression involving both, handle its component operations in a clear order, as in the third example.
Common Mistake: Adding Limits for a Product
A frequent error is to treat a product as though it were a sum. For instance, if \(\lim_{x\to a}f(x)=2\) and \(\lim_{x\to a}g(x)=3\), the incorrect calculation is:
This uses the sum of the limiting values even though the functions are multiplied. The correction is to match the law to the operation:
The sum law applies to \(f(x)+g(x)\); the product law applies to \(f(x)g(x)\). Keeping that distinction explicit prevents the error.
Why These Laws Matter
Complicated expressions are often built from simpler functions using addition and multiplication. The limit laws let us analyze those constructions piece by piece instead of returning to the definition for every expression. That is why they are central to algebraic limit techniques.
Before applying a law, ask three questions: What operation joins the functions? Do the relevant component limits exist? Are those limits finite? Once these checks are made, the calculation is usually short. The justification behind the rules also explains why they are dependable: closeness of the component values controls the closeness of their sum or product.
Check Your Understanding
Use the sum and product limit laws, and state which law you apply.
- If \(\lim_{x\to 1}f(x)=5\) and \(\lim_{x\to 1}g(x)=-2\), what is \(\lim_{x\to 1}(f(x)+g(x))\)?
- If \(\lim_{x\to 3}p(x)=-4\) and \(\lim_{x\to 3}q(x)=2\), what is \(\lim_{x\to 3}p(x)q(x)\)?
- If \(\lim_{x\to a}r(x)=6\) and \(\lim_{x\to a}s(x)=1\), find \(\lim_{x\to a}\bigl(r(x)s(x)+s(x)\bigr)\).
- In the product-law proof, why is it useful to establish that \(f(x)\) is bounded near \(a\)?
- What is wrong with using the product law when one of the stated component limits does not exist as a finite real number?
