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

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

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

Factoring gives ((x - 1)(x^3 + x^2 + x + 1))/(x - 1). For x ≠ 1, this simplifies to x^3 + x^2 + x + 1. Thus, lim (x -> 1) = 4.

Factoring gives (x - 2)(x + 5)/(x - 2). For x ≠ 2, this simplifies to x + 5. Evaluating at x = 2 gives 7.

The expression is undefined at x=2. The limit does not exist as the function approaches infinity.

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.

The expression can be factored as ((x - 3)(x + 3))/(x - 3). For x ≠ 3, this simplifies to x + 3. Thus, lim (x -> 3) (x + 3) = 6.

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

As x approaches infinity, the leading terms dominate. Thus, lim (x -> ∞) (3x^2)/(5x^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 = -b/(2a) = -4/(2*-1) = 2. f(2) = -2^2 + 4(2) + 1 = 5.

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

f'(x) = -3x^2 + 6x. Setting f'(x) = 0 gives x(3x - 6) = 0, so x = 0 or x = 2. f(2) = 5.

Set f'(x) = 0 to find critical points. The local maximum occurs at x = 2. f(2) = 5.

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

Magnitude = √(3^2 + 4^2) = √(9 + 16) = √25 = 5.

|A| = √(3^2 + (-4)^2) = √(9 + 16) = √25 = 5.

Magnitude |v| = √(3^2 + (-4)^2 + 12^2) = √(9 + 16 + 144) = √169 = 13.

Area = 1/2 * base * height. Max area occurs when height is maximized, thus Area = 1/2 * 10 * 10 = 50.

Area = 1/2 * base * height. Max area occurs when height is maximized at 10 units, giving Area = 50.

Area = 1/2 * base * height. Max area occurs when height is maximized, which is 10 units, giving Area = 50.

For maximum area, the triangle should be equilateral. Area = (sqrt(3)/4) * (10)^2 = 75 cm².

The maximum occurs at t = -b/(2a) = -32/(2*-16) = 1. h(1) = 64.

The maximum occurs at t = -b/(2a) = 64/(2*16) = 2. h(2) = -16(2^2) + 64(2) + 48 = 80.

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.