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Constant Functions

Core Definition

A constant function is a function of the form f(x)=cf(x) = c, where c∈Rc \in \mathbb{R} is a fixed real number. The function assigns the exact same output cc to every input xx in its domain. Its limit as xx approaches any value aa is simply the constant itself:

lim⁡x→ac=c\lim_{x \to a} c = c

Concept Explanation

Graphically, a constant function forms a perfectly horizontal line at height y=cy = c. Because the function's height never changes:

  1. As xx approaches any coordinate aa from the left (x→a−x \to a^-), the height is cc.
  2. As xx approaches aa from the right (x→a+x \to a^+), the height is cc.
  3. Both directional paths are constant and agree everywhere, meaning the limit is always determinate and continuous on the entire real line R\mathbb{R}.
        y ^
          |
        c +------------------- y = c
          |         |
          +---------+---------> x
                    a

Why We Use It in Calculus

  • Baseline for Limit Laws: Serves as the building block for the Constant Multiple Law and polynomial term evaluation.
  • Piecewise Boundaries: Many engineering piecewise models use constant values to represent inactive states, baseline voltages, or reset thresholds (e.g. f(x)=0f(x) = 0 when x<0x < 0).
  • Derivatives & Rates of Change: Because constant functions do not change height, their rate of change (derivative) is identically zero: ddx[c]=0\frac{d}{dx}[c] = 0.

Example Problems

Example Problem 1: Basic Evaluation of Constant Limits

Evaluate the following limits: (a) lim⁡x→517\lim_{x \to 5} 17
(b) lim⁡x→−3π\lim_{x \to -3} \pi
(c) lim⁡x→∞(−4.5)\lim_{x \to \infty} (-4.5)

Solution

Since the expression inside each limit does not depend on xx, apply the constant rule lim⁡x→ac=c\lim_{x \to a} c = c:

  • (a) lim⁡x→517=17\lim_{x \to 5} 17 = 17
  • (b) lim⁡x→−3π=π\lim_{x \to -3} \pi = \pi
  • (c) lim⁡x→∞(−4.5)=−4.5\lim_{x \to \infty} (-4.5) = -4.5

Example Problem 2: Constant Branch in a Piecewise Function

Consider the piecewise function:

f(x)={4:x<2x2:x≥2f(x) = \begin{cases} 4 & : x < 2 \\ x^2 & : x \ge 2 \end{cases}
Evaluate lim⁡x→2−f(x)\lim_{x \to 2^-} f(x) and lim⁡x→2+f(x)\lim_{x \to 2^+} f(x), and determine whether lim⁡x→2f(x)\lim_{x \to 2} f(x) exists.

Solution

Step 1: Evaluate the left-sided limit (x<2x < 2)
Approaching 22 from the left uses the constant branch:

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

Step 2: Evaluate the right-sided limit (x>2x > 2)
Approaching 22 from the right uses the quadratic branch:

lim⁡x→2+f(x)=lim⁡x→2+(x2)=22=4\lim_{x \to 2^+} f(x) = \lim_{x \to 2^+} (x^2) = 2^2 = 4

Step 3: Compare directional limits
Since lim⁡x→2−f(x)=4=lim⁡x→2+f(x)\lim_{x \to 2^-} f(x) = 4 = \lim_{x \to 2^+} f(x), by the Existence Theorem, the overall limit exists:

lim⁡x→2f(x)=4\lim_{x \to 2} f(x) = 4

Example Problem 3: Reference Voltage Baseline (Electrical Engineering)

A digital logic gate outputs a constant high voltage level Vhigh(t)=5.0 VV_{\text{high}}(t) = 5.0\text{ V} during standby mode. What is the limit of the logic gate output as time tt approaches the clock transition threshold t→10 nst \to 10\text{ ns}?

Solution

Since the gate remains at a steady DC state prior to the clock pulse:

lim⁡t→10Vhigh(t)=lim⁡t→10(5.0 V)=5.0 V\lim_{t \to 10} V_{\text{high}}(t) = \lim_{t \to 10} (5.0\text{ V}) = 5.0\text{ V}

Answer: 5.0 V5.0\text{ V}


Connections & References