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Limit Laws and Algebraic Methods · Tutorial 191 of 1495

Product Law for Limits

Use the product law to find limits of products and identify when multiplying function values at the target gives a different answer.

Beginner 8 min read

What You'll Learn

  • State the conditions required for the product law for limits.
  • Evaluate products by finding and multiplying the individual limits.
  • Apply the law when one or both functions are not defined at the target.
  • Distinguish a product’s limit from the product’s value at the target.
  • Recognize when the product law cannot be used because a finite factor limit is missing.

Why Products Have a Limit Rule

When two quantities each approach a finite value, it is natural to expect their product to approach the product of those values. The product law for limits makes that expectation precise. It lets us evaluate a product by considering the limiting behavior of its factors separately, rather than trying to analyze the product from scratch.

The rule is about limits, not necessarily about the values of the functions at the target. A function can be undefined at the target, or its value there can differ from the value it approaches. We will keep those two ideas separate throughout.

The Product Law

Theorem: Suppose \(f\) and \(g\) are defined at points sufficiently close to \(a\), though they need not be defined at \(a\). If the finite limits \(\lim_{x\to a} f(x)=L\) and \(\lim_{x\to a} g(x)=M\) exist, then the product \(f(x)g(x)\) is defined for \(x\) sufficiently close to \(a\), and \(\lim_{x\to a} [f(x)g(x)]=LM\).

In practical use, the procedure is direct: find the limit of each factor, check that both are finite, and multiply the results. The theorem does not require either function to have a value at \(a\); it concerns values at nearby points.

$$ \lim_{x\to a} f(x)g(x) = \left(\lim_{x\to a} f(x)\right) \left(\lim_{x\to a} g(x)\right) = LM. $$

The finite-limit condition matters. The product law does not say that every expression can be evaluated by assigning a limiting value to each factor: if a factor has no finite limit, the theorem’s hypotheses are not satisfied. The product might still have a limit, but that would need a separate argument.

Why the Rule Works

Write the difference between the product and its proposed limit by adding and subtracting \(L g(x)\). This gives an exact identity:

$$ f(x)g(x)-LM = \bigl(f(x)-L\bigr)g(x)+L\bigl(g(x)-M\bigr). $$

As \(x\) approaches \(a\), the difference \(f(x)-L\) becomes small, and \(g(x)-M\) becomes small. The remaining factor \(g(x)\) stays controlled near \(a\): because it approaches the finite number \(M\), its values are close to \(M\), and hence bounded, sufficiently near \(a\). Thus the first term becomes small, and the second does too. Their sum is the difference between the product and \(LM\), so the product approaches \(LM\).

This identity explains the mechanism behind the theorem: the product’s error is built from the errors in the individual factors, with the nearby values of a convergent factor remaining bounded. The later formal epsilon-delta proof makes the required “sufficiently close” choices precise.

Worked Examples

Worked Example: A Product of Linear Factors

Evaluate \(\lim_{x\to 2}(x+1)(2x-3)\).

First find the limit of each factor. Since \(x+1\) and \(2x-3\) approach their values at \(x=2\),

$$ \lim_{x\to 2}(x+1)=2+1=3, \qquad \lim_{x\to 2}(2x-3)=2(2)-3=1. $$

Both limits are finite, so the product law applies:

$$ \lim_{x\to 2}(x+1)(2x-3)=3\cdot 1=3. $$

Check by multiplying the factors first: \((x+1)(2x-3)=2x^2-x-3\). Substituting \(x=2\) into this polynomial gives \(2(2^2)-2-3=8-2-3=3\), matching the product-law result.

Worked Example: A Product with a Negative Target

Evaluate \(\lim_{x\to -1}(x^2+2x+4)(5-x)\).

For the first factor, substitute \(-1\) into the polynomial:

$$ \lim_{x\to -1}(x^2+2x+4) = (-1)^2+2(-1)+4 = 1-2+4 = 3. $$

For the second factor,

$$ \lim_{x\to -1}(5-x)=5-(-1)=6. $$

The two finite limits can be multiplied:

$$ \lim_{x\to -1}(x^2+2x+4)(5-x)=3\cdot 6=18. $$

As a check, the product at the target is \((1-2+4)(5+1)=3\cdot 6=18\). Here direct substitution agrees with the product law because both factors are polynomials and are defined at the target.

Worked Example: The Product Limit Can Differ from the Product Value

Define \(f(x)=x+2\) for \(x\ne 1\), with \(f(1)=10\), and define \(g(x)=2x\) for \(x\ne 1\), with \(g(1)=-4\). Find the limit of \(f(x)g(x)\) as \(x\to 1\), and compare it with the value of the product function at \(1\).

For \(x\ne 1\), the formulas give \(f(x)=x+2\) and \(g(x)=2x\). Therefore, as \(x\) approaches \(1\),

$$ \lim_{x\to 1} f(x)=1+2=3, \qquad \lim_{x\to 1} g(x)=2(1)=2. $$

Both limits are finite. The product law gives

$$ \lim_{x\to 1} f(x)g(x)=3\cdot 2=6. $$

But the value of the product function at \(1\) uses the specified function values there:

$$ f(1)g(1)=10\cdot(-4)=-40. $$

Thus \(\lim_{x\to 1} f(x)g(x)=6\), whereas \(f(1)g(1)=-40\). The limit describes the behavior at nearby points; it need not equal the product’s value at the target.

Check the nearby product directly: for \(x\ne 1\), \(f(x)g(x)=(x+2)(2x)=2x^2+4x\). Its limit at \(1\) is \(2(1)^2+4(1)=2+4=6\), confirming the product-law result.

Limit of a Product Versus Product at the Target

For a function \(h\), the limit \(\lim_{x\to a}h(x)\) is determined by the values of \(h(x)\) near \(a\). The value \(h(a)\), if it exists, is a separate piece of information. Applied to a product, this means:

$$ \lim_{x\to a} f(x)g(x) = \left(\lim_{x\to a}f(x)\right) \left(\lim_{x\to a}g(x)\right), $$

provided the two finite limits exist. This does not assert that the limit equals \(f(a)g(a)\). That equality does hold when the functions are defined at \(a\) and their values there agree with their respective limits. The example above shows why the distinction matters when that agreement fails.

It is also useful to distinguish multiplying limits from multiplying function values at a nearby input. For each nearby \(x\), the product function is indeed formed as \(f(x)g(x)\). The product law says that as those paired values change with \(x\), their product approaches the product \(LM\) of the two limiting values. It does not replace the nearby values with the values at \(a\).

A Common Mistake: Using the Rule Without Its Hypotheses

A frequent error is to see one factor approaching zero and conclude that the product must approach zero, even when the other factor does not have a finite limit. For \(x\ne 0\), consider \(f(x)=x\) and \(g(x)=1/x\). A mistaken use of the product law might look like this:

$$ \lim_{x\to 0} f(x)=0, \qquad \lim_{x\to 0} g(x)\text{ does not exist as a finite number}, \qquad \text{so } \lim_{x\to 0} f(x)g(x)=0. $$

The conclusion does not follow: the second factor has no finite limit, so the product law cannot be applied. In fact, for every \(x\ne 0\),

$$ f(x)g(x)=x\cdot\frac{1}{x}=1, \qquad \text{and therefore } \lim_{x\to 0}f(x)g(x)=1. $$

The correction is to check the hypotheses before multiplying limiting values. If one factor lacks a finite limit, analyze the product itself by an appropriate separate method; do not treat the missing limit as an ordinary number.

Using the Product Law Reliably

1
Identify the factors.
Write the expression as \(f(x)g(x)\) and note the target \(a\).
2
Find each factor limit.
Determine whether \(\lim_{x\to a}f(x)\) and \(\lim_{x\to a}g(x)\) exist as finite numbers.
3
Apply the rule only when justified.
If both finite limits exist, multiply them to obtain the product limit.
4
Check what the question asks.
A limit and a function value at the target are not automatically equal.

The product law is useful because it reduces one limit calculation to two simpler ones. Its conditions also prevent a tempting but invalid shortcut: multiplying limiting values is justified only when both factors have finite limits. Keeping that qualification in view makes the rule both efficient and reliable.

Check Your Understanding

For each question, state any condition needed before applying the product law.

  1. If \(\lim_{x\to a}f(x)=-3\) and \(\lim_{x\to a}g(x)=4\), what is \(\lim_{x\to a}f(x)g(x)\)?
  2. If \(f\) and \(g\) have finite limits at \(a\), must \(f(a)g(a)\) equal \(\lim_{x\to a}f(x)g(x)\)? Explain.
  3. Suppose \(f(x)=x\) and \(g(x)=1/x\) for \(x\ne0\). Why can the product law not be used to evaluate their product limit at \(0\)?
  4. If both factors are defined at \(a\) and their values equal their limits, how can you evaluate the limit of their product?
  5. In the identity \(f(x)g(x)-LM=(f(x)-L)g(x)+L(g(x)-M)\), which terms reflect the errors in the two factors?

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