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Power and Root Law

Core Definition

If lim⁡x→af(x)=L\lim_{x \to a} f(x) = L (where LL is a finite real number) and n,mn, m are positive integers (with m≠0m \neq 0), then:

lim⁡x→a[f(x)]n/m=[lim⁡x→af(x)]n/m=Ln/m\lim_{x \to a} [f(x)]^{n/m} = \left[\lim_{x \to a} f(x)\right]^{n/m} = L^{n/m}
For integer powers nn: lim⁡x→a[f(x)]n=[lim⁡x→af(x)]n=Ln\lim_{x \to a} [f(x)]^n = \left[\lim_{x \to a} f(x)\right]^n = L^n
For roots ⋅n\sqrt[n]{\cdot}: lim⁡x→af(x)n=lim⁡x→af(x)n=Ln\lim_{x \to a} \sqrt[n]{f(x)} = \sqrt[n]{\lim_{x \to a} f(x)} = \sqrt[n]{L} (provided L>0L > 0 if nn is even).

Concept Explanation

The Power and Root Law allows moving the limit operator inside exponential powers and radical signs. This property stems directly from the continuity of power and radical functions on their domains.

Crucial Constraints & Domain Restrictions

  1. Even Roots and Negative Numbers: If nn is an even integer (such as a square root ⋅\sqrt{\cdot} or 4th root ⋅4\sqrt[4]{\cdot}), we must have L>0L > 0 for the real limit to exist. If L<0L < 0, the root yields an imaginary number, so the limit does not exist in the real numbers.
  2. Fractional Powers: A rational exponent an/ma^{n/m} can be evaluated as (am)n(\sqrt[m]{a})^n or anm\sqrt[m]{a^n}. For negative bases a<0a < 0, taking the odd root first avoids intermediate negative values under even roots.

Why We Use It in Calculus

  • Radical Functions: Evaluates square roots, cube roots, and rational exponents directly by computing the limit of the radicand (inner expression) first.
  • Physical Power Scaling: Common in kinetic energy (Ek=12mv2E_k = \frac{1}{2}mv^2), fluid pressure head, and stress-strain non-linear power models.

Example Problems

Example Problem 1: Fractional Exponent with Negative Base

Evaluate the limit:

lim⁡x→0(x−8)2/3\lim_{x \to 0} (x - 8)^{2/3}

Solution

Step 1: Move the limit inside the power

lim⁡x→0(x−8)2/3=[lim⁡x→0(x−8)]2/3\lim_{x \to 0} (x - 8)^{2/3} = \left[\lim_{x \to 0} (x - 8)\right]^{2/3}

Step 2: Evaluate the inner limit

lim⁡x→0(x−8)=0−8=−8\lim_{x \to 0} (x - 8) = 0 - 8 = -8

Step 3: Simplify the fractional exponent
Rewrite (−8)2/3(-8)^{2/3} as (−83)2(\sqrt[3]{-8})^2:

  • Cube root: −83=−2\sqrt[3]{-8} = -2
  • Square: (−2)2=4(-2)^2 = 4

Answer: 44

Example Problem 2: Limit Inside a Radical

Evaluate the limit:

lim⁡x→52x2−1\lim_{x \to 5} \sqrt{2x^2 - 1}

Solution

Step 1: Pass the limit operator inside the square root

lim⁡x→52x2−1=lim⁡x→5(2x2−1)\lim_{x \to 5} \sqrt{2x^2 - 1} = \sqrt{\lim_{x \to 5} (2x^2 - 1)}

Step 2: Evaluate the radicand

lim⁡x→5(2x2−1)=2(5)2−1=2(25)−1=50−1=49\lim_{x \to 5} (2x^2 - 1) = 2(5)^2 - 1 = 2(25) - 1 = 50 - 1 = 49

Step 3: Compute the square root

49=7\sqrt{49} = 7

Answer: 77

Example Problem 3: Kinetic Energy Telemetry (Mechanical & Aerospace Engineering)

The kinetic energy of a decelerating spacecraft probe is modeled by K(t)=12m[v(t)]2K(t) = \frac{1}{2} m [v(t)]^2, where m=400 kgm = 400\text{ kg}. Doppler telemetry indicates that as the probe approaches reentry interface time (t→tct \to t_c), its velocity limit is lim⁡t→tcv(t)=30 m/s\lim_{t \to t_c} v(t) = 30\text{ m/s}. Find lim⁡t→tcK(t)\lim_{t \to t_c} K(t).

Solution

Step 1: Apply the Constant Multiple and Power Laws

lim⁡t→tcK(t)=12m⋅[lim⁡t→tcv(t)]2\lim_{t \to t_c} K(t) = \frac{1}{2} m \cdot \left[\lim_{t \to t_c} v(t)\right]^2

Step 2: Substitute values

12(400 kg)⋅[30 m/s]2=200⋅900=180,000 J=180 kJ\frac{1}{2} (400\text{ kg}) \cdot [30\text{ m/s}]^2 = 200 \cdot 900 = 180{,}000\text{ J} = 180\text{ kJ}

Answer: 180 kJ180\text{ kJ}


Connections & References