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 requires finding the horizontal asymptote of q(x) = (2x² - 1)/(x³ + 3x) as x approaches infinity. The numerator has degree 2 while the denominator has degree 3, meaning the denominator grows faster than the numerator. When the denominator's degree exceeds the numerator's degree by any amount, the horizontal asymptote is y = 0. As x approaches infinity, the function behaves like 2x²/x³ = 2/x, which approaches 0. Choice B (y = 2) incorrectly ignores the degree difference. Remember: higher denominator degree always yields horizontal asymptote y = 0.

This problem tests limits at infinity involving radicals: s(x) = √(x² + 9)/x as x approaches infinity. To evaluate this, we factor x² from inside the radical: √(x²(1 + 9/x²))/x = |x|√(1 + 9/x²)/x. For x > 0 (approaching positive infinity), |x| = x, so we get x√(1 + 9/x²)/x = √(1 + 9/x²). As x approaches infinity, 9/x² approaches 0, so the limit is √1 = 1. Choice A (0) incorrectly assumes the radical grows slower than x. When radicals contain polynomials, factor out the highest power to find limits at infinity.

This question asks for the horizontal asymptote of v(x) = (2x³ - 8)/(x³ + 4x) as x approaches negative infinity. Both numerator and denominator have degree 3, so we find the asymptote by dividing the leading coefficients: 2/1 = 2. As x approaches negative infinity, the highest-degree terms dominate, making the function behave like 2x³/x³ = 2. The horizontal asymptote is the same whether x approaches positive or negative infinity for rational functions with equal degrees. Choice D (y = 0) would only be correct if the denominator had higher degree. For equal-degree rational functions, the horizontal asymptote is always the ratio of leading coefficients.

This question tests evaluating the limit of p(x) = (4x² + 9)/(x² - 16) as x approaches infinity. Since both numerator and denominator have degree 2, we find the limit by taking the ratio of leading coefficients: 4/1 = 4. As x becomes very large, the constant terms become negligible, and the function behaves like 4x²/x² = 4. The vertical asymptotes at x = ±4 (where the denominator equals zero) don't affect the horizontal behavior at infinity. Choice B (0) would only be correct if the denominator had higher degree. For equal-degree rational functions, the limit at infinity equals the ratio of leading coefficients.

This problem requires evaluating the limit of u(x) = (ln x)/x as x approaches infinity. This is a classic limit where logarithmic growth is compared to linear growth. Since ln x grows much slower than x (logarithmic vs. linear growth), the fraction approaches 0 as x approaches infinity. This can be proven using L'Hôpital's rule: taking derivatives gives us 1/x divided by 1, which equals 1/x, confirming the limit is 0. Choice A (1) incorrectly assumes ln x and x grow at the same rate. Remember that logarithmic functions grow slower than any positive power of x.

This problem involves finding limits at infinity to determine horizontal asymptotes. For h(x) = (7 - 2x³)/(5x³ + x), the degrees are both 3, so the horizontal asymptote is the ratio of leading coefficients, -2/5, but we must consider the direction as x approaches negative infinity. Substituting x = -t where t approaches infinity yields (7 + 2t³)/(-5t³ - t), which simplifies to -2/5 for large t. Thus, the horizontal asymptote is y = -2/5. A tempting distractor is y = 2/5, which ignores the sign change due to approaching negative infinity. To find horizontal asymptotes of rational functions, compare the degrees of the numerator and denominator and, if equal, divide the leading coefficients, accounting for signs in limits to negative 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 u(x) = (6x²)/(-3x² + 10), the degrees are both 2, so the horizontal asymptote is y = 6/(-3) = -2. Even degree ensures consistency from both sides. Lower terms are negligible. A tempting distractor is y = 2, forgetting the negative sign. 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 the rational function f(x) = (5x² - 3)/(2x² + 7x + 1), the degrees of the numerator and denominator are both 2, so the horizontal asymptote is given by the ratio of the leading coefficients, which is 5/2. As x approaches infinity, the lower-degree terms become negligible, leaving the function behaving like 5x²/2x². Therefore, the limit as x approaches infinity is 5/2, corresponding to the horizontal asymptote y = 5/2. A tempting distractor is y = 2/5, which would result from incorrectly swapping the leading coefficients. To find horizontal asymptotes of rational functions, compare the degrees of the numerator and denominator and, if equal, divide the leading coefficients.