This question involves finding the horizontal asymptote of t(x) = $e^x$$/(e^x$ + 5) as x approaches infinity. As x approaches infinity, $e^x$ grows exponentially and dominates the constant 5 in the denominator. We can divide both numerator and denominator by $e^x$ to get 1/(1 + $5/e^x$). Since $e^x$ approaches infinity, $5/e^x$ approaches 0, giving us a limit of 1/(1 + 0) = 1. Therefore, the horizontal asymptote is y = 1. Choice A (y = 0) would be correct if x approached negative infinity instead. For exponential functions, identify which terms dominate as x approaches infinity.

This problem involves finding limits at infinity to determine horizontal asymptotes. For F(x) = (3x² + 7)/(6x² - 5), degrees are both 2, so the horizontal asymptote is y = 3/6 = 1/2. Even degree maintains the ratio. Negligible lower terms. A tempting distractor is y = 2, inverting the fraction. To find horizontal asymptotes of rational functions, compare the degrees of the numerator and denominator and, if equal, divide the leading coefficients.

This problem involves finding limits at infinity to determine horizontal asymptotes. For J(x) = (2x + 1)/(x - 3), degrees both 1, horizontal asymptote is y = 2/1 = 2. Odd degree but same limit from both sides. Leading terms dominate. A tempting distractor is y = 1/2, inverting the ratio. To find horizontal asymptotes of rational functions, compare the degrees of the numerator and denominator and, if equal, divide the leading coefficients.

This problem involves finding limits at infinity to determine horizontal asymptotes for logarithmic functions. As x approaches infinity, ln(x) grows much slower than x, so v(x) = ln(x)/x approaches 0. This can be confirmed using L'Hôpital's rule since it's ∞/∞ form: derivative is (1/x)/1 = 1/x → 0. Thus, the limit is 0. A tempting distractor like ∞ might arise if someone thinks ln(x) grows faster than x, which is incorrect. A transferable strategy for horizontal asymptotes with logs and polynomials is to recognize that logarithms grow slower than any positive power of x, leading to a limit of zero when divided by x.

This question assesses understanding of limits at infinity, particularly for logarithmic over polynomial growth. For r(x) = ln(x)/x, as x approaches infinity, ln(x) grows slower than any positive power of x, including x itself. Therefore, the ratio approaches 0 by known limit properties or L'Hôpital's rule, since it's ∞/∞ form: derivative is (1/x)/1 = 1/x → 0. This means the function approaches the x-axis. A tempting distractor could be infinity, possibly confusing the growth rates of log and linear functions. In general, compare growth rates of functions to determine limits at infinity, using tools like L'Hôpital for indeterminate forms.

This question assesses understanding of limits at infinity, for rational functions as x approaches negative infinity. For t(x) = (x² - 4x + 1)/(2x² + 7), divide by x²: (1 - 4/x + 1/x²)/(2 + 7/x²). As x approaches negative infinity, the fractional terms approach zero, yielding 1/2. The sign consistency arises because both leading terms are positive and degrees match. A tempting distractor could be -1/2, maybe from incorrectly introducing a negative sign for negative infinity. In general, for equal-degree rationals, the asymptote is the leading coefficient ratio, independent of infinity's direction.

This question assesses understanding of limits at infinity, for functions involving square roots. For s(x) = √(x² + 9)/x, factor x² out of the square root: √(x²(1 + 9/x²))/x = |x| √(1 + 9/x²)/x. As x approaches positive infinity, |x|/x = 1, and √(1 + 9/x²) → 1, so the limit is 1, giving y=1 as the asymptote. Note that for x → -∞, it would be -1, but the question specifies x → ∞. A tempting distractor might be y=-1, perhaps overlooking the direction of infinity. In general, simplify expressions by factoring the dominant term inside roots or powers to find horizontal asymptotes.

This problem involves finding limits at infinity to determine horizontal asymptotes for rational functions. To find the horizontal asymptote of f(x) as x approaches infinity, divide both the numerator and denominator by x², the highest power in the denominator, yielding (3 - 5/x + 1/x²)/(2 + 7/x²). As x approaches infinity, the terms with 1/x and 1/x² approach zero, so the limit simplifies to 3/2. Therefore, the horizontal asymptote is y = 3/2. A tempting distractor like y = 2/3 might arise if someone mistakenly swaps the leading coefficients of the numerator and denominator. A transferable strategy for horizontal asymptotes in rational functions is to compare the degrees of the numerator and denominator; if equal, the asymptote is the ratio of the leading coefficients.

This problem tests limits at infinity involving square roots, requiring careful algebraic manipulation. For s(x) = √(x² + 4x)/x, we need to factor out x² from under the radical: √(x²(1 + 4/x))/x = |x|√(1 + 4/x)/x. As x → ∞, x is positive so |x| = x, giving us x√(1 + 4/x)/x = √(1 + 4/x). As x approaches infinity, 4/x approaches 0, so the limit is √(1 + 0) = 1. Students often incorrectly think the answer is 0 or infinity without properly simplifying the radical expression. When dealing with radicals in limits at infinity, always factor out the highest power of x from under the radical first.

This question asks for the limit at infinity of a rational function, which determines its horizontal asymptote. For g(x) = (4x - 9)/(x + 3), the numerator has degree 1 and the denominator has degree 1. When degrees are equal, we find the limit by dividing the leading coefficients: 4/1 = 4. As x approaches infinity, the constant terms -9 and +3 become negligible compared to the x terms. Students might incorrectly choose 0 (choice B) thinking the degree in the denominator is higher, or -3 (choice C) by looking at the denominator's constant. Remember: for rational functions with equal degrees in numerator and denominator, the horizontal asymptote is the ratio of leading coefficients.