Continuity

Core Definition

A function $f(x)$ is continuous at a specific point $x = c$ if and only if all three of the following conditions are satisfied without exception:

$f(c)$ is defined: $c$ sits within the domain of $f$ , producing a real, finite value.
$\lim_{x \to c} f(x)$ exists: The left-hand and right-hand paths converge to the identical real number: $\lim_{x \to c^-} f(x) = \lim_{x \to c^+} f(x) = L$
$\lim_{x \to c} f(x) = f(c)$ : The approached limit equals the exact evaluated point value.

A function is continuous on an interval if it is continuous at every interior point and continuous from the appropriate side at any endpoints.

Concept Explanation

The three conditions form a hierarchical testing procedure:

Condition 1 (Existence of Target Point): Checks whether the function is physically defined at $c$ . If $c$ causes division by zero, a negative inside an even root, or logarithm of zero/negative, Condition 1 fails immediately.
Condition 2 (Existence of Trajectory): Checks whether the graph approaches a unified destination from both directions. If the paths separate (a jump) or shoot to infinity (a vertical asymptote), Condition 2 fails.
Condition 3 (Agreement of Trajectory and Target): Checks whether the trajectory actually meets the plotted point. If there is a hole at the curve with the dot plotted elsewhere, Condition 3 fails.

If $f$ and $g$ are continuous at $x = c$ , then:

$f \pm g$ , $c \cdot f$ , and $f \cdot g$ are continuous at $c$ .
$\frac{f}{g}$ is continuous at $c$ provided $g(c) \neq 0$ .
Composition of Continuous Functions: If $g$ is continuous at $c$ and $f$ is continuous at $g(c)$ , then the composite function $(f \circ g)(x) = f(g(x))$ is continuous at $c$ (see Limits of Composite Functions).

Structural Integrity & Stress Fields (Civil & Mechanical Engineering): Stress fields across welded structural joints must be continuous. A discontinuity in shear or bending moment indicates an unmodeled concentrated crack or sudden material interface failure.
Microcontroller Signal Smoothing (Computer & Electrical Engineering): Analog sensor inputs passing into digital feedback control loops are tested for continuity to prevent abrupt actuator chatter or sudden motor torque surges.
Thermodynamic Phase Boundaries (Chemical & Materials Engineering): Thermal and chemical phase changes (such as solid-to-liquid transitions) produce discontinuities in enthalpy and heat capacity, requiring piecewise modeling to program automated reactor cooling systems safely.

Determine whether $f(x)$ is continuous at $x = 1$ :

f(x) = \begin{cases} 2x + 3 & : x < 1 \\ 5 & : x = 1 \\ x^2 + 4 & : x > 1 \end{cases}

Solution

Condition 1: Check $f(1)$ : Using the explicit point definition at $x = 1$ :

f(1) = 5 \quad \text{(Defined)}

Condition 2: Check $\lim_{x \to 1} f(x)$ :

Since $\lim_{x \to 1^-} f(x) = \lim_{x \to 1^+} f(x) = 5$ , the overall limit exists:

\lim_{x \to 1} f(x) = 5

Condition 3: Check agreement:

\lim_{x \to 1} f(x) = 5 = f(1)

Conclusion: All three criteria are satisfied, so $f(x)$ is continuous at $x = 1$ .

Find the value of $k$ that makes $g(x)$ continuous everywhere:

g(x) = \begin{cases} \frac{x^2 - 9}{x - 3} & : x \neq 3 \\ kx - 4 & : x = 3 \end{cases}

Solution

Step 1: Compute the limit as $x \to 3$ : Direct substitution gives $\frac{0}{0}$ . Factoring the numerator via Factor and Cancel:

\lim_{x \to 3} \frac{(x - 3)(x + 3)}{x - 3} = \lim_{x \to 3} (x + 3) = 3 + 3 = 6

Step 2: Evaluate the function value at $x = 3$ :

g(3) = k(3) - 4 = 3k - 4

Step 3: Force Condition 3 ( $\lim_{x \to 3} g(x) = g(3)$ ):

6 = 3k - 4 \implies 3k = 10 \implies k = \frac{10}{3}

Conclusion: Setting $k = \frac{10}{3}$ eliminates the removable discontinuity and makes $g(x)$ continuous everywhere.

Parent Concepts: Limits, Existence Theorem for Two-Sided Limits
Direct Theoretical Extensions: Discontinuities, Limits of Composite Functions, Squeeze Theorem, Solving Inequalities
Course Links: Calc-1 Session 02 | Calc-1 Session 03 | Calc-1 Session 05