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.

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.

Continuity is crucial because it allows us to apply many theorems and techniques, such as the Intermediate Value Theorem and the Mean Value Theorem. It ensures predictable behavior of functions.

Each condition plays a specific role. f(c) being defined ensures there's a value at the point. The limit existing ensures the function approaches a single value. The limit equaling f(c) ensures there's no jump or hole.

For a function to be continuous at a point, the limit at that point must exist. This means the left-hand limit and the right-hand limit must be equal.

Visually, a function is continuous if its graph can be drawn without lifting your pen. There should be no jumps, breaks, or holes in the graph at the point in question.

If a function is differentiable at a point, it must also be continuous at that point. However, the converse is not always true; a function can be continuous but not differentiable (e.g., at a sharp corner).

Rational functions are discontinuous where the denominator is equal to zero, because division by zero is undefined. This creates a vertical asymptote or a hole in the graph.

The Intermediate Value Theorem states that if a function is continuous on a closed interval [a, b], then for any value k between f(a) and f(b), there exists a c in [a, b] such that f(c) = k. Continuity is a requirement for this theorem.

One-sided limits (left-hand and right-hand limits) are crucial for determining continuity at endpoints of intervals or at points where the function's definition changes. For continuity, both one-sided limits must exist and be equal to the function's value at that point.

Continuity is used in physics to model motion, in engineering to design structures, and in economics to analyze market trends. It ensures that models behave predictably and realistically.

Piecewise functions are defined by different expressions on different intervals. Continuity must be checked at the points where the function definition changes to ensure the pieces connect smoothly.