🌱 Tony's Memex Tony's Intertwingled Memex
Digital Garden | OER Curriculum
Home / Digital Garden / Sum Law

Sum Law

🖨️ Printable Reference / PDF
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 sum is the sum of the individual limits.

Concept Explanation

The Sum Law establishes the additive linearity of the limit operator. It allows you to evaluate the limit of a combined additive expression by decomposing it into independent, simpler limit calculations.

Crucial Constraints & Conditions

  1. Finite Existence Requirement: The Sum Law is valid if and only if both individual component limits exist as finite real numbers.
  2. Indeterminate Form Caution: If one or both component limits yield ±∞\pm\infty or an indeterminate form like 00\frac{0}{0}, you cannot separate the terms using the Sum Law. You must first combine or simplify the expression algebraically before taking the limit.

Why We Use It in Calculus

  • Polynomial Evaluation: Enables term-by-term evaluation of any polynomial or algebraic sum without requiring numerical tables or graph approximations.
  • Superposition of Subsystems: Allows engineers to evaluate composite multi-variable inputs by analyzing the asymptotic behavior of each signal or physical component independently.

Example Problems

Example Problem 1: Evaluating from Known Component Limits

Given that lim⁡x→2f(x)=5\lim_{x \to 2} f(x) = 5 and lim⁡x→2g(x)=−8\lim_{x \to 2} g(x) = -8, evaluate:

lim⁡x→2[f(x)+g(x)]\lim_{x \to 2} [f(x) + g(x)]

Solution

Step 1: Verify conditions
Both lim⁡x→2f(x)=5\lim_{x \to 2} f(x) = 5 and lim⁡x→2g(x)=−8\lim_{x \to 2} g(x) = -8 exist and are finite real numbers. The Sum Law applies directly.

Step 2: Apply the Sum Law

lim⁡x→2[f(x)+g(x)]=lim⁡x→2f(x)+lim⁡x→2g(x)\lim_{x \to 2} [f(x) + g(x)] = \lim_{x \to 2} f(x) + \lim_{x \to 2} g(x)

Step 3: Substitute and compute

5+(−8)=−35 + (-8) = -3

Answer: −3-3

Example Problem 2: Term-by-Term Polynomial Decomposition

Evaluate the limit analytically using the Sum Law:

lim⁡x→3(x2+4x+1)\lim_{x \to 3} (x^2 + 4x + 1)

Solution

Step 1: Distribute the limit across the sum

lim⁡x→3(x2+4x+1)=lim⁡x→3x2+lim⁡x→34x+lim⁡x→31\lim_{x \to 3} (x^2 + 4x + 1) = \lim_{x \to 3} x^2 + \lim_{x \to 3} 4x + \lim_{x \to 3} 1

Step 2: Evaluate each term

  • lim⁡x→3x2=32=9\lim_{x \to 3} x^2 = 3^2 = 9
  • lim⁡x→34x=4(3)=12\lim_{x \to 3} 4x = 4(3) = 12
  • lim⁡x→31=1\lim_{x \to 3} 1 = 1

Step 3: Combine results

9+12+1=229 + 12 + 1 = 22

Answer: 2222

Example Problem 3: Multi-Branch Sensor Telemetry (Electrical Engineering)

An electrical engineer measures the combined current Itotal(t)=I1(t)+I2(t)I_{\text{total}}(t) = I_1(t) + I_2(t) at a circuit junction during system startup (t→0 mst \to 0\text{ ms}). Sensor telemetry shows lim⁡t→0I1(t)=3.5 A\lim_{t \to 0} I_1(t) = 3.5\text{ A} and lim⁡t→0I2(t)=2.1 A\lim_{t \to 0} I_2(t) = 2.1\text{ A}. Compute lim⁡t→0Itotal(t)\lim_{t \to 0} I_{\text{total}}(t).

Solution

Step 1: Apply the Sum Law to the circuit current model

lim⁡t→0Itotal(t)=lim⁡t→0[I1(t)+I2(t)]=lim⁡t→0I1(t)+lim⁡t→0I2(t)\lim_{t \to 0} I_{\text{total}}(t) = \lim_{t \to 0} [I_1(t) + I_2(t)] = \lim_{t \to 0} I_1(t) + \lim_{t \to 0} I_2(t)

Step 2: Substitute sensor values

3.5 A+2.1 A=5.6 A3.5\text{ A} + 2.1\text{ A} = 5.6\text{ A}

Answer: 5.6 A5.6\text{ A}

Connections & References