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

The function is not continuous at x = 0 since the limit does not equal f(0).

To check continuity at x = 1, we find the left limit (5) and the right limit (2). They are not equal, hence f(x) is not continuous at x = 1.

For continuity at x = 1, the left limit (3) must equal f(1) (2), which is not true.

For continuity at x = 1, both pieces must equal 4, hence the function is continuous.

To check continuity at x = 1, we find the left limit (3) and the right limit (3). Both equal 3, hence f(x) is continuous at x = 1.

Limit as x approaches 0 from left is 0, and f(0) = 1, hence it is not continuous at x = 0.

To check continuity at x = 0, we find f(0) = 1 and limit as x approaches 0 is also 1.

The function is not continuous at x = 1 because the left-hand limit does not equal the function value.

To check continuity at x = 1, we find f(1) = 1, limit as x approaches 1 from left is 1, and from right is also 1.

To check continuity at x = 1, we find the limit from both sides. Both limits equal 1, hence f(x) is continuous at x = 1.

Both conditions must hold true for continuity at x = 2.

To be continuous at x = 2, k must equal f(2) = 2^2 = 4.

As x approaches 0 from the right, f(x) approaches infinity, indicating a discontinuity at x = 0.

Setting 2k + 2 = 0 gives k = 2.

Setting k(0) = 0^2 + 1 gives k = 1.

Setting k(2) = 2^2 gives 2k = 4, thus k = 2.

Setting the two pieces equal at x = 2: 3(2) + p = 2^2 - 4. Solving gives p = -2.

Setting the two pieces equal at x = 1: 5 - q = 3(1) + 2. Solving gives q = 0.

Setting 2(1) + 1 = a and a = 2 for continuity.

Setting 2(3) + a = 5 gives a = -1.

Setting 4 = 2 gives a = 1 for continuity.

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

Setting the left limit (3(2) + 1 = 7) equal to the right limit (2^2 - 1 = 3), we find m = 3.

f(x) = x^2 - 4 is a polynomial function and is continuous everywhere, including at x = 2.

At x = 2, f(2) = 0 and limit from left is 0, limit from right is also 0. Hence, it is continuous.

To check continuity at x = 2, we find the left limit (4), right limit (4), and f(2) (4). All are equal, so f(x) is continuous.

f(x) = x^2 is a polynomial function and is continuous everywhere.

The function f(x) = 1/x is not defined at x = 0, hence it is not continuous there.

The function has a jump discontinuity at x = 1, hence it is not continuous.