\(\lim_{x\to a^{-}} f(x) = \lim_{x\to a^{+}} f(x) = f(a))

Check if (\lim_{x\to c^{-}} f(x) = \lim_{x\to c^{+}} f(x) = f(c))

\(\sqrt{z}) requires (z \geq 0)

\(\frac{1}{z}) requires (z \neq 0)

\(f(x) = a_n x^n + a_{n-1} x^{n-1} + ... + a_1 x + a_0)

Solve the inequality (g(x) \geq 0).

Solve the equation (g(x) = 0) and exclude those values from the domain.

If (f) is continuous on ([a, b]) and (k) is a number between (f(a)) and (f(b)), then there exists at least one (c) in ([a, b]) such that (f(c) = k).

If (\lim_{x \to a} f(x)) exists but is not equal to (f(a)), or (f(a)) is undefined, then there is a removable discontinuity at (x = a).

If (f(x)) is differentiable at (x=a), then (f(x)) is continuous at (x=a).

Step 1: Find the left-hand limit. Step 2: Find the right-hand limit. Step 3: Evaluate the function at that point. Step 4: Compare all three values; they must be equal for continuity.

Step 1: Identify values where (q(x) = 0). Step 2: Check if these values are within the interval. Step 3: If any are, the function is discontinuous over the interval.

Step 1: Set the expression inside the square root greater than or equal to zero: (g(x) \geq 0). Step 2: Solve the inequality for (x). Step 3: The solution is the domain of the function.

Step 1: Set the pieces of the function equal to each other at the point where the definition changes. Step 2: Solve for 'k'. Step 3: Verify that this value of 'k' makes the function continuous by checking the left-hand and right-hand limits.

Step 1: Check for any domain restrictions within the interval. Step 2: If there are no domain restrictions, the function is likely continuous. Step 3: For piecewise functions, check continuity at each transition point.

Step 1: Simplify the function: (f(x) = x + 2) for (x \neq 2). Step 2: Check the limit as (x \to 2): (\lim_{x \to 2} (x + 2) = 4). Step 3: Since the limit exists but (f(2)) is undefined, it's a removable discontinuity.

Step 1: Show (f(x)) is continuous on ((a,b)). Step 2: Show (\lim_{x \to a^+} f(x) = f(a)). Step 3: Show (\lim_{x \to b^-} f(x) = f(b)).

Step 1: Solve (x^2 - 5x + 6 \geq 0). Step 2: Factor the quadratic: ((x - 2)(x - 3) \geq 0). Step 3: Determine the intervals where the inequality holds: ((-\infty, 2] \cup [3, \infty)).

Step 1: Find the left-hand limit: (\lim_{x \to 1^-} x^2 = 1). Step 2: Find the right-hand limit: (\lim_{x \to 1^+} (2x - 1) = 1). Step 3: Evaluate (f(1) = 2(1) - 1 = 1). Step 4: Since all three values are equal, the function is continuous at (x = 1).

Step 1: Set the two pieces equal at (x = 2): (c(2) + 2 = (2)^2 - c(2)). Step 2: Solve for 'c': (2c + 2 = 4 - 2c \Rightarrow 4c = 2 \Rightarrow c = \frac{1}{2}).

Polynomials are continuous everywhere because they have no breaks, jumps, or vertical asymptotes in their graphs.

Rational functions are continuous everywhere except where the denominator is zero, which creates a vertical asymptote or a hole.

You must verify that the left-hand limit, right-hand limit, and the function's value at that point are all equal to ensure there are no jumps or breaks.

Domain restrictions indicate points where the function is not defined, leading to discontinuities. These points must be excluded when determining continuity over an interval.

For a function to be continuous at a point, the limit as x approaches that point must exist and be equal to the function's value at that point.

A removable discontinuity occurs when a function has a hole at a point, meaning the limit exists, but the function is either undefined or has a different value at that point. It can be 'removed' by redefining the function at that point.

A non-removable discontinuity occurs when the limit does not exist at a point, such as at a vertical asymptote or a jump discontinuity. It cannot be 'removed' by redefining the function at that point.

Checking both limits is essential for determining continuity, especially in piecewise functions, because the function's behavior may differ on either side of a point.

Trigonometric functions like sine and cosine are continuous everywhere. Tangent, cotangent, secant, and cosecant are continuous on their domains, excluding points where they have vertical asymptotes.

Exponential functions are continuous everywhere. Logarithmic functions are continuous on their domains, which are typically (x > 0).