Using the limit property, lim (x -> 0) (sin(kx)/x) = k. Here, k = 5, so the limit is 5, and 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.

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 lim (x -> 0) (sin(x)/x) = 1. Since the limit exists and equals the function value at x = 0, it is continuous.

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.

To find critical points, set f'(x) = 0. f'(x) = 3x^2 - 6x = 3x(x - 2). Critical points are x = 0 and x = 2.

Using the power rule, f'(x) = 12x^2 - 2.

Using the power rule, f'(x) = 10x + 3.

Using the power rule, f'(x) = 10x + 3.

Using the power rule, f'(x) = 15x^2 - 4.

Using the product rule, f'(x) = (x^3)' * ln(x) + x^3 * (ln(x))' = 3x^2 * ln(x) + x^2.

Using the power rule, f'(x) = 4x^3 + 6x^2 - 1.

Using the power rule, f'(x) = 4x^3 - 12x^2 + 12x.

Using the power rule, f'(x) = 4x^3 - 12x^2 + 12x - 24.

Using the power rule, f'(x) = 5x^4 - 6x^2 + 1.

Using the derivatives of sine and cosine, g'(x) = cos(x) - sin(x).

Using the fact that sin(x) approaches x as x approaches 0, we have lim (x -> 0) (x^3)/(sin(x)) = 0.

As x approaches infinity, the leading terms dominate. Thus, lim (x -> ∞) (3x^2)/(5x^2) = 3/5.

f'(x) = -2x + 6; setting to 0 gives x = 3; f(3) = -3^2 + 6(3) - 8 = 1.

The vertex occurs at x = 2. f(2) = 2^2 - 4*2 + 5 = 1, so local minima is (2, 1).

The maximum occurs at x = -b/(2a) = 10/(2*2) = 2.5. f(2.5) = -2(2.5^2) + 10(2.5) - 12 = 6.

The maximum occurs at x = -b/(2a) = -12/(-6) = 2. f(2) = -3(2^2) + 12(2) - 5 = 7.

The vertex is at x = -4/(2*(-1)) = 2. The maximum value is f(2) = -2^2 + 4*2 + 5 = 7.

The minimum occurs at x = -b/(2a) = 16/(2*4) = 2. f(2) = 4(2^2) - 16(2) + 15 = 1.

The vertex is at x = 8/(2*4) = 1. The minimum value is f(1) = 4(1)^2 - 8(1) + 3 = -1.

The minimum occurs at x = -b/(2a) = -6/(2*1) = -3. f(-3) = (-3)^2 + 6(-3) + 10 = 1.

The minimum occurs at x = -b/(2a) = 12/(2*3) = 2. f(2) = 3(2^2) - 12(2) + 9 = 3.

First derivative f'(x) = 16x^3 - 6x^2 + 1. Second derivative f''(x) = 48x^2 - 12x.