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.

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

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

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.

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

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

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

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

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

At x = 0, lim x→0- f(x) = 0 and lim x→0+ f(x) = 1, hence it is discontinuous at x = 0.

The function f(x) = 1/(x-3) is not defined at x = 3, hence it is not continuous at that point.

The function f(x) = |x| is continuous everywhere, including at x = 0.