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Difference Law

Formulas & Definitions
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) - g(x)] = \lim_{x \to a} f(x) - \lim_{x \to a} g(x) = L - M
The limit of a difference is the difference of the individual limits.

Concept Explanation

The Difference Law is a direct mathematical consequence of combining the Sum Law with the Constant Multiple Law (with scalar c=−1c = -1):

lim⁡x→a[f(x)−g(x)]=lim⁡x→a[f(x)+(−1)g(x)]=lim⁡x→af(x)−lim⁡x→ag(x)\lim_{x \to a} [f(x) - g(x)] = \lim_{x \to a} [f(x) + (-1)g(x)] = \lim_{x \to a} f(x) - \lim_{x \to a} g(x)

Crucial Constraints & Conditions

  1. Finite Component Limits: Both lim⁡x→af(x)\lim_{x \to a} f(x) and lim⁡x→ag(x)\lim_{x \to a} g(x) must evaluate to finite numbers.
  2. Indeterminate Form ∞−∞\infty - \infty: If evaluating each term individually produces ∞−∞\infty - \infty, the Difference Law cannot be applied directly. This is an indeterminate form that requires algebraic simplification (e.g., finding a common denominator or multiplying by a conjugate) before taking the limit.

Why We Use It in Calculus

Example Problems

Example Problem 1: Basic Evaluation with Negative Values

Given that lim⁡x→−1f(x)=−4\lim_{x \to -1} f(x) = -4 and lim⁡x→−1g(x)=7\lim_{x \to -1} g(x) = 7, evaluate:

lim⁡x→−1[f(x)−g(x)]\lim_{x \to -1} [f(x) - g(x)]

Solution

Step 1: Check conditions
Both limits exist and are finite real numbers.

Step 2: Apply the Difference Law

lim⁡x→−1[f(x)−g(x)]=lim⁡x→−1f(x)−lim⁡x→−1g(x)\lim_{x \to -1} [f(x) - g(x)] = \lim_{x \to -1} f(x) - \lim_{x \to -1} g(x)

Step 3: Substitute values

−4−7=−11-4 - 7 = -11

Answer: −11-11

Example Problem 2: Radical and Linear Difference

Evaluate the limit:

lim⁡x→4(x−2x)\lim_{x \to 4} (\sqrt{x} - 2x)

Solution

Step 1: Apply the Difference Law

lim⁡x→4(x−2x)=lim⁡x→4x−lim⁡x→42x\lim_{x \to 4} (\sqrt{x} - 2x) = \lim_{x \to 4} \sqrt{x} - \lim_{x \to 4} 2x

Step 2: Evaluate individual limits

  • lim⁡x→4x=4=2\lim_{x \to 4} \sqrt{x} = \sqrt{4} = 2
  • lim⁡x→42x=2(4)=8\lim_{x \to 4} 2x = 2(4) = 8

Step 3: Compute final difference

2−8=−62 - 8 = -6

Answer: −6-6

Example Problem 3: Voltage Drop Analysis (Electrical Engineering)

An electronics engineer calculates the net voltage drop across a filter network: Vdrop(t)=Vin(t)−Vout(t)V_{\text{drop}}(t) = V_{\text{in}}(t) - V_{\text{out}}(t). As operating temperature stabilizes (T→25∘CT \to 25^\circ\text{C}), telemetry confirms lim⁡T→25Vin(T)=12.0 V\lim_{T \to 25} V_{\text{in}}(T) = 12.0\text{ V} and lim⁡T→25Vout(T)=4.8 V\lim_{T \to 25} V_{\text{out}}(T) = 4.8\text{ V}. Find lim⁡T→25Vdrop(T)\lim_{T \to 25} V_{\text{drop}}(T).

Solution

Step 1: Distribute the limit using Difference Law

lim⁡T→25Vdrop(T)=lim⁡T→25Vin(T)−lim⁡T→25Vout(T)\lim_{T \to 25} V_{\text{drop}}(T) = \lim_{T \to 25} V_{\text{in}}(T) - \lim_{T \to 25} V_{\text{out}}(T)

Step 2: Substitute telemetry values

12.0 V−4.8 V=7.2 V12.0\text{ V} - 4.8\text{ V} = 7.2\text{ V}

Answer: 7.2 V7.2\text{ V}

Connections & References