The left limit is 0 and the right limit is undefined. Thus, f(x) is not continuous at x = 0.

At x = 1, f(1) = 3, but lim x->1- f(x) = 1 and lim x->1+ f(x) = 2. Thus, it is not continuous.

Both sides equal 2 at x = 1, hence it is continuous.

The limit as x approaches 0 does not exist, hence f(x) is not continuous at x = 0.

The left limit is 0, the right limit is 2, and f(1) = 3. Thus, it is discontinuous.

The left limit as x approaches 1 is 1, and the right limit is also 1. Thus, f(1) = 1, making it continuous.

Setting the two pieces equal at x = 1 gives us 3 + c = 1. Thus, c = -2.

To ensure continuity at x = 1, k(1) + 1 = 2(1) - 3, solving gives k = 2.

To ensure continuity at x = 1, we need to set the two pieces equal: 1^2 + k = 2(1) + 1. This gives k = 2.

To ensure continuity at x = 1, we need to set the two pieces equal: k + 1^2 = 2(1) + 3. This gives k + 1 = 5, so k = 4.

For f(x) to be continuous at x = 2, we need limit as x approaches 2 from left to equal limit as x approaches 2 from right and equal to f(2). Thus, k = 4.

For f(x) to be continuous at x = 2, we need limit as x approaches 2 from left to equal limit as x approaches 2 from right. Thus, k must equal 0.

Setting the two pieces equal at x = 0: n^2 - 1 = 3. Solving gives n = ±2.

Setting 2(2) + 3 = p(2) + 1 gives p = 3 for continuity.

Setting 3(2) - 1 = p(2) + 4 gives p = 2.

Setting p = 1 for continuity at x = 0 gives f(0) = 1.

Setting the left limit (1 - 1 = 0) equal to the right limit (2(1) + 1 = 3), we find p = 2.

Setting -3 + p = 3 gives p = 0.

Setting 1 - 3 + p = 2 + 1 gives p = 4.

Factoring gives (x-1)(x^2 + x + 1)/(x - 1). Canceling (x - 1) gives lim x->1 (x^2 + x + 1) = 3.

Factoring gives (x-1)(x^2 + x + 1)/(x-1) = x^2 + x + 1, thus limit is 3.

Factoring gives (x-2)(x+2)/(x-2). Canceling gives lim x->2 (x + 2) = 4.

Factoring gives (x-2)(x+2)/(x-2) = x + 2, thus limit is 4.

Using the limit property, lim x->0 (sin(kx)/x) = k. Here, k = 3, so the limit is 3.

Using L'Hôpital's rule, the limit evaluates to 3.

Setting ax + 1 = 2 and x^2 + a = 2 at x = 1 gives a = 0.

Setting ax + 1 = 3 and 2x + a = 3 at x = 1 gives a = 2.

Setting the two pieces equal at x = 2 gives us 2a + 1 = 1. Solving for a gives a = 0.

Setting the two pieces equal at x = 2: 2a + 1 = 2^2 - 3. Solving gives a = 2.

Setting the left limit (1 + a) equal to the right limit (3), we find a = 2.