Continuity: Function has no breaks, jumps, or holes. | Differentiability: Function has a derivative at every point (smooth, no sharp corners or vertical tangents).

Removable: Limit exists, but doesn't equal f(c) or f(c) is undefined. | Jump: Left and right limits exist, but are not equal.

Continuous: Satisfies all three conditions for continuity at every point in its domain. | Discontinuous: Fails at least one of the three conditions for continuity at one or more points in its domain.

Left-hand limit: The value f(x) approaches as x approaches c from the left (x < c). | Right-hand limit: The value f(x) approaches as x approaches c from the right (x > c).

Continuity at a point: Function satisfies the three conditions for continuity at a specific x-value. | Continuity on an interval: Function is continuous at every point within that interval.

Rational function: Can have discontinuities where the denominator is zero. | Polynomial function: Continuous everywhere.

Vertical asymptote: Represents an infinite discontinuity where the function approaches infinity. | Hole: Represents a removable discontinuity where the function is undefined but the limit exists.

Direct substitution: Plug in the value directly into the function. Works for continuous functions. | Limit laws: Apply algebraic rules to simplify and evaluate limits, especially when direct substitution fails.

Intermediate Value Theorem: Guarantees a value between f(a) and f(b) for a continuous function on [a, b]. | Extreme Value Theorem: Guarantees a maximum and minimum value for a continuous function on [a, b].

Definition: Rigorous approach using limits and function values. | Graph: Visual approach identifying breaks, jumps, or holes.

1. Check if f(1) is defined: $f(1) = 1^2 + 2(1) + 1 = 4$. 2) Find $\lim_{x \to 1} f(x)$: $\lim_{x \to 1} (x^2 + 2x + 1) = 4$. 3) Check if $\lim_{x \to 1} f(x) = f(1)$: $4 = 4$. Since all conditions are met, f(x) is continuous at x = 1.

1. Check if f(2) is defined: $f(2) = \frac{1}{2-2} = \frac{1}{0}$, which is undefined. Since f(2) is undefined, f(x) is discontinuous at x = 2.

1. Find the left-hand limit: $\lim_{x \to 2^-} (x + 1) = 3$. 2) Find the right-hand limit: $\lim_{x \to 2^+} (kx) = 2k$. 3) Set the limits equal: $3 = 2k$. 4) Solve for k: $k = \frac{3}{2}$.

1. Check if f(0) is defined: $f(0) = |0| = 0$. 2) Find $\lim_{x \to 0^-} f(x) = 0$ and $\lim_{x \to 0^+} f(x) = 0$. Therefore, $\lim_{x \to 0} f(x) = 0$. 3) Check if $\lim_{x \to 0} f(x) = f(0)$: $0 = 0$. Since all conditions are met, f(x) is continuous at x = 0.

1. Check if f(1) is defined: $f(1) = \frac{1^2 - 1}{1 - 1} = \frac{0}{0}$, which is undefined. 2) Simplify the function: $f(x) = \frac{(x-1)(x+1)}{x-1} = x + 1$ for $x \neq 1$. 3) Find $\lim_{x \to 1} f(x) = \lim_{x \to 1} (x+1) = 2$. Since f(1) is undefined, f(x) is discontinuous at x = 1 (removable discontinuity).

1. Evaluate the function at the point from both sides (left and right). 2) Ensure the left-hand limit equals the right-hand limit. 3) Verify that the value of the function at that point matches the limit.

Show that the function is continuous at every point within the interval. For closed intervals, also show continuity at the endpoints (using one-sided limits).

Set the denominator of the rational function equal to zero and solve for x. These x-values represent the points of discontinuity.

Polynomial functions are continuous everywhere because the limit as x approaches any value c is equal to the function's value at c: $\lim_{x \to c} P(x) = P(c)$.

If the limit exists at the point but is not equal to the function's value at that point, or if the function is undefined at that point, then the function has a removable discontinuity.

Look for jumps, holes (open circles), or vertical asymptotes. These indicate points where the function is not continuous.

A smooth, unbroken curve indicates that the function is continuous over that interval. There are no sudden jumps or breaks.

A vertical asymptote indicates an infinite discontinuity. The function is not continuous at the x-value where the asymptote occurs.

An open circle indicates a removable discontinuity (a hole). The function is not defined at that specific x-value, or the limit does not equal the function value.

A jump in the graph indicates a jump discontinuity. The left-hand limit and the right-hand limit exist but are not equal.

Check if the pieces of the graph connect smoothly at the points where the function definition changes. There should be no gaps or jumps.

If a function is differentiable, its graph is smooth and continuous. There are no sharp corners, cusps, or discontinuities.

The graph of $f(x) = \frac{1}{x}$ has a vertical asymptote at x = 0, indicating a discontinuity at that point. The function is continuous everywhere else.

The graph of $f(x) = |x|$ is continuous everywhere. Although it has a sharp corner at x = 0, it is still continuous at that point.

Trace the graph from both the left and right sides towards the x-value of interest. If the y-values approach the same number, that number is the limit.