Using the limit property lim (x -> 0) (tan(kx)/x) = k, we have lim (x -> 0) (tan(5x)/x) = 5.

Factoring gives (x - 1)(x^3 + x^2 + x + 1)/(x - 1). Canceling gives lim (x -> 1) (x^3 + x^2 + x + 1) = 4.

Dividing the numerator and denominator by x^2, we get lim (x -> ∞) (5 + 3/x^2)/(2 + 1/x^2) = 5/2.

Using the standard limit lim (x -> 0) (tan(kx)/x) = k, we have k = 5. Thus, lim (x -> 0) (tan(5x)/x) = 5.

Factoring gives (x - 1)(x^2 + x + 1)/(x - 1). For x ≠ 1, this simplifies to x^2 + x + 1. Evaluating at x = 1 gives 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. Evaluating at x = 1 gives 4.

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.

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

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

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.

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 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.