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

Dividing numerator and denominator by x^2 gives lim (x -> ∞) (3 + 2/x^2)/(5 - 4/x^2) = 3/5.

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.

f'(x) = -2x + 4. Setting f'(x) = 0 gives x = 2. Since f''(x) = -2 < 0, it is a maxima.

f'(x) = 2x; thus, f'(3) = 2(3) = 6.

Using the product rule, f'(x) = 4e^x + 4x^2e^x.

Using the chain rule, f'(x) = 2x/(x^2 + 1).

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

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.

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

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

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

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

First derivative f'(x) = 3x^2 - 6x; second derivative f''(x) = 6x - 6.

f'(x) = cos(x) - sin(x). At x = π/4, f'(π/4) = cos(π/4) - sin(π/4) = √2/2 - √2/2 = 0.

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.

The second derivative f''(x) = d^2/dx^2(e^x) = e^x.

f'(x) = e^x and f''(x) = e^x. Therefore, f''(0) = e^0 = 1.

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.

First, find f'(x) = 3x^2 - 6x. Then, f'(2) = 3(2^2) - 6(2) = 12 - 12 = 0.

Find f''(x) = 6x - 12. Set f''(x) = 0 gives x = 2. The inflection point is (2, f(2)) = (2, 0).

f'(x) = 4x^3 - 12x^2; thus, f'(2) = 4(2^3) - 12(2^2) = 32 - 48 = -16.

Using the chain rule, h'(x) = 2e^(2x).

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.