∫_1^2 (3x^2 - 2) dx = [x^3 - 2x] from 1 to 2 = (8 - 4) - (1 - 2) = 3.

The integral of 1/(x^2 + 1) is tan^(-1)(x) + C.

Integrating term by term gives (1/4)x^4 - (1/3)x^3 + 4x + C.

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.

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

Dividing by x^2, lim(x→∞) (3 + 2/x^2)/(5 - 4/x^2) = 3/5.

Using the identity 1 - cos(x) = 2sin^2(x/2), we have lim (x -> 0) (2sin^2(x/2))/(x^2) = 1/2.

Using L'Hôpital's Rule, we differentiate the numerator and denominator: lim (x -> 0) (e^x)/(1) = e^0 = 1.

Using L'Hôpital's Rule, differentiate the numerator and denominator: lim (x -> 0) (1/(1 + x))/(1) = 1.

Using the standard limit lim (x -> 0) (sin(x)/x) = 1, we have lim (x -> 0) (sin(5x)/x) = 5 * lim (x -> 0) (sin(5x)/(5x)) = 5 * 1 = 5.

Using the standard limit lim (x -> 0) (tan(x)/x) = 1, we have lim (x -> 0) (tan(3x)/x) = 3 * lim (x -> 0) (tan(3x)/(3x)) = 3 * 1 = 3.

Using the limit property lim (x -> 0) (tan(x)/x) = 1, we find that the limit is 1.

Using the Taylor series expansion for sin(x), we find that lim (x -> 0) (x - sin(x))/x^3 = 1/6.

Since |sin(1/x)| <= 1, we have |x^2 * sin(1/x)| <= x^2, and thus lim (x -> 0) x^2 * sin(1/x) = 0.

As x approaches 0, sin(x) approaches x, thus lim (x -> 0) (x^2)/(sin(x)) = 0.

As x approaches 0, x^3 approaches 0 and sin(x) approaches 0, thus the limit is 0.

Using L'Hôpital's Rule, the limit evaluates to 2.

Factoring gives (x - 3)(x + 3)/(x - 3). Canceling (x - 3) gives lim (x -> 3) (x + 3) = 6.

Divide numerator and denominator by x^2. The limit becomes lim (x -> ∞) (2 + 3/x^2)/(5 - 4/x + 1/x^2) = 2/5.

Dividing numerator and denominator by x^3 gives lim (x -> ∞) (2 - 3/x^2)/(4 + 5/x^3) = 2/4 = 1/2.

Dividing numerator and denominator by x^2 gives lim (x -> ∞) (3 + 0)/(5 - 0) = 3/5.