It involves redefining the function at the point of discontinuity to equal the limit at that point, effectively 'filling the hole' and making the function continuous.

Factoring allows you to simplify rational functions and cancel out common factors that cause discontinuities, revealing the simplified function and the location of the hole.

The left-hand limit and the right-hand limit at the boundary must exist and be equal to each other, and the function's value at the boundary must also be equal to this limit.

Because the limit exists at that point, and by redefining the function at that point, we can make the function continuous.

The graph appears continuous except for a single point (a 'hole') where the function is undefined.

Because the function's behavior may be different on either side of the boundary point, and continuity requires these limits to match.

Factor the numerator and denominator, cancel any common factors, and identify the x-values that were canceled out.

The limit tells us where the function 'should' be, even if it isn't defined there. This allows us to redefine the function and remove the discontinuity.

Graphs provide a visual representation of where a function might have breaks, jumps, or holes, making it easier to spot potential discontinuities.

Setting the denominator equal to zero helps to find potential points of discontinuity, and setting the numerator equal to zero helps to find the x-intercepts of the function.

Factor the numerator: $f(x) = \frac{(x - 2)(x + 2)}{x - 2}$. Cancel common factors: $f(x) = x + 2$. Redefine $f(2) = 4$.

Set $x^2 = ax$ at $x = 1$. Solve for $a$: $1^2 = a(1)$, so $a = 1$.

Factor: $f(x) = \frac{(x - 1)(x + 1)}{x + 1}$. Cancel: $f(x) = x - 1$. Yes, at $x = -1$.

Factor: $f(x) = \frac{(x - 3)(x + 3)}{x - 3}$. Cancel: $f(x) = x + 3$. Define $f(3) = 6$.

Set $x + a = x^2$ at $x = 0$. Solve for $a$: $0 + a = 0^2$, so $a = 0$.

Factor the numerator: $f(x) = \frac{(x - 1)(x - 3)}{x - 1}$. Cancel the common factor: $f(x) = x - 3$. Evaluate at x = 1: $f(1) = 1 - 3 = -2$.

Set the two pieces equal at x = 2: $c(2)^2 = 2(2) + c$. Solve for c: $4c = 4 + c$, so $3c = 4$, and $c = \frac{4}{3}$.

Factor the numerator: $f(x) = \frac{(x - 5)(x + 5)}{x + 5}$. Cancel the common factor: $f(x) = x - 5$. Evaluate at x = -5: $f(-5) = -5 - 5 = -10$.

Factor the numerator: $f(x) = \frac{(x - 3)(x + 3)}{x - 3}$. Cancel the common factor: $f(x) = x + 3$. Evaluate at x = 3: $f(3) = 3 + 3 = 6$. Set k = 6.

Factor the numerator: $f(x) = \frac{(x + 1)(x + 2)}{x + 2}$. Cancel the common factor: $f(x) = x + 1$. Yes, it has a removable discontinuity at x = -2. Evaluate at x = -2: $f(-2) = -2 + 1 = -1$.

Removable: Limit exists, can be 'fixed'. Non-removable: Limit doesn't exist, cannot be 'fixed'.

Jump: Function 'jumps' from one value to another. Infinite: Function approaches infinity (vertical asymptote).

Direct Substitution: Simple, works if function is continuous. Factoring: Used when direct substitution leads to indeterminate form.

Continuous: No breaks, holes, or jumps. Discontinuous: Has breaks, holes, jumps, or asymptotes.

Graphically: Visual estimation of where the function is heading. Algebraically: Using techniques like factoring or rationalizing to find the exact value.

At a point: Limit exists and equals the function's value. Over an interval: Continuous at every point in the interval.

Hole: Removable discontinuity where the limit exists. Vertical Asymptote: Non-removable discontinuity where the function approaches infinity.

Rational Functions: Look for zeros in the denominator. Piecewise Functions: Check the boundaries between the pieces.

Intermediate Value Theorem: Guarantees a function takes on every value between any two given points if continuous. Extreme Value Theorem: Guarantees the function has a max and min if continuous.

Direct Substitution: First attempt, works if not indeterminate. L'Hopital's Rule: Used for indeterminate forms like 0/0 or ∞/∞, involves derivatives.