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 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.

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.