The left limit as x approaches 1 is 1, the right limit is 2, and f(1) = 2. Since the left and right limits do not match, f(x) is not continuous at x = 1.

At x = 1, f(1) = 2(1) - 1 = 1 and lim x→1- f(x) = 1, lim x→1+ f(x) = 1. Thus, f(x) is continuous at x = 1.

The function f(x) = |x| is continuous at x = 0 since both the left-hand limit and right-hand limit equal f(0) = 0.

Using the limit property, lim (x -> 0) (sin(5x)/x) = 5. The function is continuous at x = 0.

Using the limit property, lim (x -> 0) (sin(5x)/x) = 5. The function is continuous at x = 0 if defined as f(0) = 5.

The limit is 1, and if we define f(0) = 1, then f(x) is continuous at x = 0.

lim (x -> 3) (x^2 - 9)/(x - 3) = lim (x -> 3) (x + 3) = 6. The function is not defined at x = 3, hence not continuous.

Using L'Hôpital's Rule, lim x→2 (x^2 - 4)/(x - 2) = lim x→2 (2x)/(1) = 4.

f(x) is a polynomial function and is continuous everywhere, hence no points of discontinuity.

The left limit as x approaches 0 is 0, the right limit is 1, and f(0) = 0. Since the limits do not match, f(x) is discontinuous at x = 0.

To make f(x) continuous at x = 1, we need c = 1^2 = 1. Thus, c must be 1.

The function f(x) = kx + 2 is a linear function and is continuous for all k at x = 3.

To ensure continuity at x = 2, k(2) + 1 must equal 3. Thus, k = 1.

f(1) = 3(1) + 2 = 5. Since f(x) is a linear function, it is continuous everywhere.

f(2) = 3(2) + 2 = 8. Since f(x) is a polynomial, it is continuous everywhere.

f(-1) = (-1)^2 + 2*(-1) + 1 = 0. The function is a polynomial and is continuous everywhere, including at x = -1.

lim x→-1 f(x) = (-1)^2 + 3(-1) + 2 = 1 - 3 + 2 = 0.

f(-1) = (-1)^2 + 3(-1) + 2 = 0. Since f(x) is a polynomial, it is continuous everywhere.

At x = 1, f(1) = 1^2 = 1 and the limit from the left is also 1, hence f(x) is continuous at x = 1.

At x = 1, f(1) = 3 and limit from left is 1^2 = 1. Since they are not equal, f(x) is discontinuous at x = 1.

f(1) = 1^3 - 3*1 + 2 = 0. The function is a polynomial and hence continuous everywhere, including at x = 1.

f(1) = 1^3 - 3(1) + 2 = 0. Since f(x) is a polynomial, it is continuous everywhere.

The function f(x) = 1/(x-1) is not defined at x = 1, hence it is discontinuous at that point.

The function f(x) = sqrt(x) is continuous at x = 0 as it is defined and the limit exists.

The function f(x) = |x| is continuous at x = 0 because the left limit, right limit, and f(0) all equal 0.

The function f(x) = 1/(x-1) is discontinuous at x = 1, hence it is continuous on (1, ∞).

f(x) = x^2 + 3 is a polynomial function and is continuous for all x in (-∞, ∞).

The function f(x) = x^2 is a polynomial function and is continuous at all points, including -1, 0, and 1.

The function f(x) = x^3 - 3x is a polynomial function and is continuous at all points, including -2, 0, and 2.

The left limit as x approaches 0 is 0, but the right limit is 1. Hence, it is not continuous at x = 0.