A function f(x) is continuous at x=c if: 1) f(c) is defined, 2) $\lim_{x \to c} f(x)$ exists, and 3) $\lim_{x \to c} f(x) = f(c)$.

A function is discontinuous at x=c if at least one of the three conditions for continuity is not met: f(c) is undefined, $\lim_{x \to c} f(x)$ does not exist, or $\lim_{x \to c} f(x) \neq f(c)$.

The left-hand limit is the value that f(x) approaches as x approaches c from values less than c, denoted as $\lim_{x \to c^-} f(x)$.

The right-hand limit is the value that f(x) approaches as x approaches c from values greater than c, denoted as $\lim_{x \to c^+} f(x)$.

A removable discontinuity occurs when $\lim_{x \to c} f(x)$ exists, but either f(c) is undefined or $\lim_{x \to c} f(x) \neq f(c)$. It can be 'removed' by redefining f(c).

A jump discontinuity occurs when the left-hand limit and the right-hand limit both exist but are not equal: $\lim_{x \to c^-} f(x) \neq \lim_{x \to c^+} f(x)$.

An infinite discontinuity occurs when the function approaches infinity (or negative infinity) as x approaches c from either the left or right, often associated with vertical asymptotes.

An oscillating discontinuity occurs when the function oscillates infinitely many times near a point, preventing the limit from existing.

The domain of a function is the set of all possible input values (x-values) for which the function is defined.

The range of a function is the set of all possible output values (y-values) that the function can produce.

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.

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.