This question tests application of rational functions to real-world contexts, specifically average cost models in economics. The function C(x) = (2000 + 50x)/x can be rewritten as C(x) = 2000/x + 50, showing that as production x increases, the fixed cost term 2000/x approaches 0. Therefore, lim[x→∞] C(x) = 0 + 50 = 50, representing the long-run average cost per item when fixed costs are spread over many units. This horizontal asymptote at y = 50 represents the variable cost per unit. Choice A correctly identifies this as $50, the long-run average cost. Choice B incorrectly suggests costs become free, misunderstanding that only the fixed cost component approaches zero. In economic applications, horizontal asymptotes often represent long-run equilibrium values.

This question tests applying rational function end behavior to a real-world context. As production x becomes very large, we evaluate lim(x→∞) A(x) = lim(x→∞) (2000 + 50x)/(x + 10). Since both numerator and denominator are linear (degree 1), we divide the leading coefficients: 50/1 = 50. This means as production increases indefinitely, the average cost per item approaches $50, which represents the variable cost per item when fixed costs become negligible. Choice A correctly identifies lim(x→∞) A(x) = 50 and HA: y = 50. Choice B incorrectly gives 200, perhaps misunderstanding the limit process. Choice C wrongly suggests the cost approaches 0, which would be economically unrealistic. In context, the horizontal asymptote represents the long-run average cost when fixed costs are spread over many items.

This question tests simplification of rational functions and identification of asymptotes after canceling common factors. The function simplifies to f(x) = (x+1)/x after canceling the common factor (x-1), creating a hole at x = 1. The vertical asymptote occurs only at x = 0 where the simplified denominator is zero. For the horizontal asymptote, the simplified function has degree 1 in both numerator and denominator with leading coefficients 1/1 = 1, so y = 1. We can verify: lim[x→∞] (x+1)/x = lim[x→∞] (1 + 1/x) = 1. Choice A correctly identifies VA: x = 0 and HA: y = 1. Choice B incorrectly identifies x = 1 as a vertical asymptote when it's actually a removable discontinuity. Always simplify first to distinguish between holes and asymptotes.

This question tests understanding of horizontal asymptotes and end behavior of rational functions using leading-term analysis. When both numerator and denominator have the same degree (both degree 2 here), the horizontal asymptote is found by dividing the leading coefficients. The leading term of the numerator is 3x² and of the denominator is x², so lim(x→∞) f(x) = 3/1 = 3. This gives a horizontal asymptote at y = 3. Choice B is correct because it accurately identifies this limit. Choice A incorrectly suggests the limit is 0, which would only occur if the numerator had lower degree. Choice C suggests the limit is infinity, which would happen if the numerator had higher degree. Choice D gives 1/3, which reverses the ratio of leading coefficients. To avoid errors, always compare degrees first: if equal, divide leading coefficients; if numerator degree is less, limit is 0; if greater, limit is ±∞.

This question tests finding asymptotes when the denominator is a perfect square. The denominator x² + 4x + 4 = (x + 2)², which equals zero only at x = -2 (with multiplicity 2). Since the numerator -2x² + 8 = -2(x² - 4) = -2(x - 2)(x + 2) doesn't have (x + 2) as a factor, there's a vertical asymptote at x = -2. For the horizontal asymptote, both numerator and denominator have degree 2, so we compare leading coefficients: lim(x→±∞) f(x) = -2/1 = -2, giving HA: y = -2. Choice A correctly identifies VA: x = -2 and HA: y = -2. Choice B incorrectly places the vertical asymptote at x = 2 instead of x = -2. Choice C gets the vertical asymptote right but incorrectly gives the horizontal asymptote as positive 2. Remember that the sign of the leading coefficient matters for horizontal asymptotes.

This question tests simplification of rational functions and identification of removable discontinuities versus vertical asymptotes. The function simplifies to f(x) = (x+1)/(x-2) after canceling the common factor (x-3), creating a hole at x = 3 rather than a vertical asymptote. The only vertical asymptote occurs at x = 2 where the simplified denominator is zero. Since the simplified function has degree 1 in both numerator and denominator, the horizontal asymptote is y = 1/1 = 1, and lim[x→∞] f(x) = 1. Choice A correctly identifies VA: x = 2, HA: y = 1, and the limit as 1. Choice B incorrectly identifies x = 3 as a vertical asymptote, missing that this is a removable discontinuity. Remember to always simplify rational functions first to distinguish between holes and vertical asymptotes.

This question tests understanding of rational functions where the numerator has higher degree than the denominator. The numerator has degree 3 while the denominator has degree 2, which means there's no horizontal asymptote. As x → ∞, the function behaves like x³/x² = x → ∞, and as x → -∞, it behaves like x → -∞. The vertical asymptotes occur where x² - 4 = 0, giving x = ±2. Choice B correctly identifies both vertical asymptotes at x = ±2, no horizontal asymptote, and the limits as ∞ and -∞. Choice A incorrectly suggests a horizontal asymptote at y = 0, which only occurs when the denominator degree exceeds the numerator degree. Remember: higher degree in numerator means no horizontal asymptote and unbounded behavior.

This question tests understanding of end behavior for rational functions where numerator and denominator have the same degree. The function has degree 2 in both numerator and denominator, so the horizontal asymptote is found by dividing leading coefficients: 3/1 = 3. As x approaches ±∞, the highest degree terms dominate the behavior, giving lim[x→±∞] (3x²/x²) = 3. The denominator x² - 2x + 1 = (x-1)² shows a vertical asymptote at x = 1. Choice B correctly states that both limits equal 3. Choice A incorrectly suggests the limits are 0, which would only occur if the denominator had higher degree than the numerator. To master this concept, always compare polynomial degrees first, then use leading coefficients for equal degrees.

This question tests understanding of rational functions, specifically finding vertical and horizontal asymptotes and evaluating end behavior limits. Vertical asymptotes occur where the denominator equals zero and the numerator doesn't, so setting x² - 4 = 0 gives x = ±2. For horizontal asymptotes, we compare the degrees of numerator and denominator polynomials - both are degree 2, so we divide the leading coefficients: 2/1 = 2, giving y = 2. As x approaches ±∞, the highest degree terms dominate, so lim[x→±∞] f(x) = lim[x→±∞] (2x²/x²) = 2. Choice A correctly identifies VA: x = ±2, HA: y = 2, and both limits approaching 2. Choice C incorrectly calculates the horizontal asymptote as 1/2, likely by confusing the coefficient ratio.

This question tests understanding of end behavior when the numerator has higher degree than the denominator, resulting in no horizontal asymptote. The numerator has degree 3 while the denominator has degree 2, so the function grows without bound. For large positive x, the dominant term ratio is x³/x² = x, which approaches +∞. For large negative x, we get x³/x² = x, which approaches -∞ since x is negative. There is no horizontal asymptote; instead, the function has oblique asymptotic behavior. Choice C correctly states lim[x→∞] f(x) = ∞ and lim[x→-∞] f(x) = -∞. Choice A incorrectly suggests horizontal asymptotic behavior at y = 1. When numerator degree exceeds denominator degree by 1, perform polynomial long division to find the oblique asymptote.