Using partial fraction decomposition, we can integrate to find that ∫ (2x + 1)/(x^2 + x) dx = ln|x^2 + x| + C.

Using the identity tan^2(x) = sec^2(x) - 1, we find that ∫ (tan(x))^2 dx = tan(x) - x + C.

The integrand simplifies to x + 1. Therefore, ∫ (x + 1) dx = (1/2)x^2 + x + C.

The integral of sin(kx) is -1/k * cos(kx). Here, k = 2, so ∫ sin(2x) dx = -cos(2x)/2 + C.

f'(x) = 4x^3 - 12x^2 = 4x^2(x - 3). Critical points are x = 0 and x = 3. Test intervals: f' is positive in (0, 3) and (3, ∞).

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.

Using L'Hôpital's rule, lim(x→0) (sin(5x)/x) = 5.

The limit is 0.

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

Using the identity 1 - cos(x) = 2sin^2(x/2), we have lim (x -> 0) (2sin^2(2x))/(x^2) = 8.

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

Using the Taylor series expansion for cos(x), we find that lim (x -> 0) (cos(x) - 1)/x^2 = -1/2.

Using the derivative of e^x at x = 0, we find lim (x -> 0) (e^x - 1)/x = 1.

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

Since |sin(1/x)| <= 1, we have |x^2 * sin(1/x)| <= |x^2|. As x approaches 0, |x^2| approaches 0, hence the limit is 0.

As x approaches 0, e^x - 1 approaches 0. Using L'Hôpital's Rule three times, we find the limit approaches 0.

This is an indeterminate form (0/0). Factor the numerator: (x-1)(x+1)/(x-1) = x + 1. Thus, lim (x -> 1) (x + 1) = 2.

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

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

The expression (x^2 - 4)/(x - 2) can be factored as (x - 2)(x + 2)/(x - 2). Canceling (x - 2) gives lim (x -> 2) (x + 2) = 4.

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

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

The maximum occurs at x = 2. f(2) = -2^2 + 4(2) + 1 = 5.

The vertex form gives maximum at x = 2. f(2) = -2^2 + 4*2 = 4.

The function is a downward-opening parabola. The maximum occurs at x = -b/(2a) = 8/(2*2) = 2. f(2) = -2(2^2) + 8(2) - 3 = 8.

The function is a downward-opening parabola. The maximum occurs at x = -b/(2a) = 8/(2*2) = 2. f(2) = -2(2^2) + 8(2) - 5 = 9.

The vertex occurs at x = 2. f(2) = -2^2 + 4(2) + 1 = 9, which is the maximum value.

The vertex occurs at x = 3. f(3) = -3^2 + 6(3) - 8 = 6.

The vertex occurs at x = -b/(2a) = 12/6 = 2. f(2) = 3(2^2) - 12(2) + 7 = 1.