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Product Law

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Core Definition

If lim⁡x→af(x)=L\lim_{x \to a} f(x) = L and lim⁡x→ag(x)=M\lim_{x \to a} g(x) = M (where LL and MM are finite real numbers), then:

lim⁡x→a[f(x)⋅g(x)]=(lim⁡x→af(x))⋅(lim⁡x→ag(x))=L⋅M\lim_{x \to a} [f(x) \cdot g(x)] = \left(\lim_{x \to a} f(x)\right) \cdot \left(\lim_{x \to a} g(x)\right) = L \cdot M
The limit of a product is the product of the individual limits.

Concept Explanation

The Product Law allows you to compute the limit of multiplied expressions by evaluating the limit of each factor separately and multiplying the results.

Crucial Constraints & Notes

  1. Finite Existence Requirement: Both factors must independently possess finite, determinate limits.
  2. Indeterminate Form 0⋅∞0 \cdot \infty: If one factor approaches 00 while the other approaches ±∞\pm\infty, the Product Law cannot be used directly. The form 0⋅∞0 \cdot \infty is indeterminate; you must rewrite the expression as a fraction (e.g. f(x)1/g(x)\frac{f(x)}{1/g(x)}) to yield 00\frac{0}{0} or ∞∞\frac{\infty}{\infty} and simplify before evaluating.

Why We Use It in Calculus

  • Higher Powers of Functions: Repeated application of the Product Law generates the Power Law for integer exponents: lim⁡[f(x)]n=[lim⁡f(x)]n\lim [f(x)]^n = [\lim f(x)]^n.
  • Coupled Physical Quantities: In engineering, many physical variables are products of distinct measured quantities, such as electrical power (P=V⋅IP = V \cdot I), mechanical work/thrust (W=F⋅dW = F \cdot d), or rocket momentum (p=m⋅vp = m \cdot v). The Product Law allows analyzing each physical factor independently.

Example Problems

Example Problem 1: Discrete Component Multiplication

Given that lim⁡x→3f(x)=6\lim_{x \to 3} f(x) = 6 and lim⁡x→3g(x)=−2\lim_{x \to 3} g(x) = -2, evaluate:

lim⁡x→3[f(x)⋅g(x)]\lim_{x \to 3} [f(x) \cdot g(x)]

Solution

Step 1: Check conditions
Both limits exist and are finite numbers (66 and −2-2).

Step 2: Apply the Product Law

lim⁡x→3[f(x)⋅g(x)]=(lim⁡x→3f(x))⋅(lim⁡x→3g(x))\lim_{x \to 3} [f(x) \cdot g(x)] = \left(\lim_{x \to 3} f(x)\right) \cdot \left(\lim_{x \to 3} g(x)\right)

Step 3: Multiply

(6)⋅(−2)=−12(6) \cdot (-2) = -12

Answer: −12-12

Example Problem 2: Product of Transcendental and Algebraic Terms

Evaluate the limit:

lim⁡x→0[(x+3)⋅cos⁡(x)]\lim_{x \to 0} [ (x + 3) \cdot \cos(x) ]

Solution

Step 1: Distribute the limit across the product

lim⁡x→0[(x+3)⋅cos⁡(x)]=(lim⁡x→0(x+3))⋅(lim⁡x→0cos⁡(x))\lim_{x \to 0} [ (x + 3) \cdot \cos(x) ] = \left(\lim_{x \to 0} (x + 3)\right) \cdot \left(\lim_{x \to 0} \cos(x)\right)

Step 2: Evaluate each limit

  • lim⁡x→0(x+3)=0+3=3\lim_{x \to 0} (x + 3) = 0 + 3 = 3
  • lim⁡x→0cos⁡(x)=cos⁡(0)=1\lim_{x \to 0} \cos(x) = \cos(0) = 1

Step 3: Multiply

3⋅1=33 \cdot 1 = 3

Answer: 33

Example Problem 3: Electrical Power Dissipation (Electrical Engineering)

Instantaneous power dissipation in a load resistor is modeled by P(t)=V(t)⋅I(t)P(t) = V(t) \cdot I(t). As an AC circuit stabilizes (t→t0t \to t_0), voltage and current telemetry readings converge to:

  • lim⁡t→t0V(t)=120 V\lim_{t \to t_0} V(t) = 120\text{ V}
  • lim⁡t→t0I(t)=2.5 A\lim_{t \to t_0} I(t) = 2.5\text{ A}

Calculate the limiting steady-state power consumption lim⁡t→t0P(t)\lim_{t \to t_0} P(t).

Solution

Step 1: Apply the Product Law

lim⁡t→t0P(t)=(lim⁡t→t0V(t))⋅(lim⁡t→t0I(t))\lim_{t \to t_0} P(t) = \left(\lim_{t \to t_0} V(t)\right) \cdot \left(\lim_{t \to t_0} I(t)\right)

Step 2: Substitute telemetry values

(120 V)⋅(2.5 A)=300 W(120\text{ V}) \cdot (2.5\text{ A}) = 300\text{ W}

Answer: 300 W300\text{ W}

Connections & References