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Constant Multiple Law

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Core Definition

If lim⁡x→af(x)=L\lim_{x \to a} f(x) = L (where LL is a finite real number) and cc is any constant real scalar (c∈Rc \in \mathbb{R}), then:

lim⁡x→a[c⋅f(x)]=c⋅lim⁡x→af(x)=c⋅L\lim_{x \to a} [c \cdot f(x)] = c \cdot \lim_{x \to a} f(x) = c \cdot L
The limit of a constant times a function is the constant times the limit of the function.

Concept Explanation

The Constant Multiple Law states that scalar constants factor cleanly out of the limit operator. Geometrically, scaling a function vertically by a factor of cc scales its approached limit value by the exact same factor cc.

Crucial Constraints & Notes

  1. cc Must Be Independent of xx: The factor cc must be a constant value that does not change as x→ax \to a. Expressions containing the variable xx cannot be factored out of the limit.
  2. Infinite Limits Caution: If lim⁡x→af(x)=±∞\lim_{x \to a} f(x) = \pm\infty, multiplying by a constant scales the magnitude, but the sign depends on whether c>0c > 0 or c<0c < 0 (and if c=0c = 0, 0⋅∞0 \cdot \infty is an indeterminate form that requires prior simplification).

Why We Use It in Calculus

  • Simplifying Algebraic Fractions: Allows pulling out numerical coefficients from numerators and denominators to isolate zero-inducing factors for cancellation.
  • Amplification and Scaling: In control systems and signal processing, scaling a sensor input by an amplifier gain KK scales the output limit directly by KK.

Example Problems

Example Problem 1: Direct Scalar Scaling

Given that lim⁡x→5f(x)=12\lim_{x \to 5} f(x) = 12, evaluate:

lim⁡x→5[−4f(x)]\lim_{x \to 5} [-4f(x)]

Solution

Step 1: Pull out the constant scalar

lim⁡x→5[−4f(x)]=−4⋅lim⁡x→5f(x)\lim_{x \to 5} [-4f(x)] = -4 \cdot \lim_{x \to 5} f(x)

Step 2: Substitute and compute

−4⋅12=−48-4 \cdot 12 = -48

Answer: −48-48

Example Problem 2: Factoring Coefficients in Indeterminate Forms

Evaluate the limit:

lim⁡x→27(x2−4)x−2\lim_{x \to 2} \frac{7(x^2 - 4)}{x - 2}

Solution

Step 1: Factor the constant multiple 77 out of the limit

lim⁡x→27(x2−4)x−2=7⋅lim⁡x→2x2−4x−2\lim_{x \to 2} \frac{7(x^2 - 4)}{x - 2} = 7 \cdot \lim_{x \to 2} \frac{x^2 - 4}{x - 2}

Step 2: Factor the remaining polynomial

x2−4=(x−2)(x+2)x^2 - 4 = (x - 2)(x + 2)
7⋅lim⁡x→2(x−2)(x+2)x−2=7⋅lim⁡x→2(x+2)7 \cdot \lim_{x \to 2} \frac{(x - 2)(x + 2)}{x - 2} = 7 \cdot \lim_{x \to 2} (x + 2)

Step 3: Evaluate the simplified limit

7⋅(2+2)=7⋅4=287 \cdot (2 + 2) = 7 \cdot 4 = 28

Answer: 2828

Example Problem 3: Signal Amplification (Electrical Engineering)

An operational amplifier has a fixed closed-loop voltage gain factor Av=−25A_v = -25. The input sensor signal voltage approaches a steady limit lim⁡t→t0Vin(t)=0.32 V\lim_{t \to t_0} V_{\text{in}}(t) = 0.32\text{ V}. Find the limiting output voltage Vout(t)=Av⋅Vin(t)V_{\text{out}}(t) = A_v \cdot V_{\text{in}}(t) as t→t0t \to t_0.

Solution

Step 1: Apply the Constant Multiple Law

lim⁡t→t0Vout(t)=lim⁡t→t0[Av⋅Vin(t)]=Av⋅lim⁡t→t0Vin(t)\lim_{t \to t_0} V_{\text{out}}(t) = \lim_{t \to t_0} [A_v \cdot V_{\text{in}}(t)] = A_v \cdot \lim_{t \to t_0} V_{\text{in}}(t)

Step 2: Substitute the gain and sensor limit

−25⋅(0.32 V)=−8.0 V-25 \cdot (0.32\text{ V}) = -8.0\text{ V}

Answer: −8.0 V-8.0\text{ V}

Connections & References