Limits at Infinity and Horizontal Asymptotes - AP Calculus BC

For $u(x)=\dfrac{2x^2+3x+5}{-4x^2+x-1}$ , what is $\lim_{x\to-\infty} u(x)$ ?
Answer: $-\dfrac{1}{2}$ This problem requires finding the limit of u(x) = (2x² + 3x + 5)/(-4x² + x - 1) as x approaches negative infinity, testing limits at infinity. Both numerator and denominator have degree 2, so we divide all terms by x² to get (2 + 3/x + 5/x²)/(-4 + 1/x - 1/x²). As x approaches negative infinity, all terms with x in the denominator approach 0, leaving us with 2/(-4) = -1/2. The horizontal asymptote is y = -1/2. Students might choose 1/2 by forgetting the negative sign in the denominator's leading coefficient. For rational functions with equal degrees, the horizontal asymptote is always the ratio of leading coefficients, including their signs.
Given $q(x)=\dfrac{2x^2+5x}{x^2-9}$ , what is the horizontal asymptote of $q$ ?
Answer: $y=2$ This problem involves finding limits at infinity to determine horizontal asymptotes. For q(x) = (2x² + 5x)/(x² - 9), the degrees are both 2, so the horizontal asymptote is y = 2/1 = 2. As x approaches infinity, the function approximates 2x²/x². This holds for both positive and negative infinity since the degree is even. A tempting distractor is y = 1/2, which swaps the coefficients incorrectly. To find horizontal asymptotes of rational functions, compare the degrees of the numerator and denominator and, if equal, divide the leading coefficients.
Given $R(x)=\dfrac{2-x}{x+8}$ , what is the horizontal asymptote as $x\to\infty$ ?
Answer: $y=-1$ This problem involves finding limits at infinity to determine horizontal asymptotes. For R(x) = (2 - x)/(x + 8), degrees both 1, horizontal asymptote as x approaches infinity is -1/1 = -1. Odd degree but same limit both sides. Leading terms. A tempting distractor is y = 1, missing negative. To find horizontal asymptotes of rational functions, compare the degrees of the numerator and denominator and, if equal, divide the leading coefficients.
Given $u(x)=\dfrac{6x^2}{-3x^2+10}$ , what is the horizontal asymptote of $u(x)$ ?
Answer: $y=-2$ 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.
For $f(x)=\dfrac{5x^2-3x+1}{2x^2+7}$ , what is the horizontal asymptote of $f$ as $x\to\infty$ ?
Answer: $y=\dfrac{5}{2}$ This problem tests finding limits at infinity for rational functions to determine horizontal asymptotes. For $f(x)=\frac{5x^2-3x+1}{2x^2+7}$ , since the numerator and denominator have the same degree (both degree 2), we find the horizontal asymptote by dividing the leading coefficients: $\frac{5}{2}$ . As $x\to\infty$ , the highest degree terms dominate, making the function behave like $\frac{5x^2}{2x^2}=\frac{5}{2}$ . The lower degree terms become negligible for large $x$ values. Choice E ( $y=\frac{2}{5}$ ) incorrectly inverts the ratio of leading coefficients. When degrees are equal in a rational function, always divide the leading coefficient of the numerator by the leading coefficient of the denominator to find the horizontal asymptote.
For $s(x)=\dfrac{3x^2+2x+5}{x^2-8x+16}$ , what is $\lim_{x\to-\infty} s(x)$ ?
Answer: $y=3$ This problem involves finding limits at infinity to determine horizontal asymptotes for rational functions. To find the limit of s(x) as x approaches negative infinity, divide by x²: (3 + 2/x + 5/x²) / (1 - 8/x + 16/x²). As x approaches negative infinity, terms with 1/x and 1/x² approach zero, yielding 3/1 = 3. Therefore, the limit is 3. A tempting distractor like y = 1/3 might result from swapping the leading coefficients. 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.
If $t(x)=\dfrac{6}{x-2}+4$ , what is the horizontal asymptote of $t$ as $x\to\infty$ ?
Answer: $y=4$ This question involves finding the horizontal asymptote of a transformed rational function. For t(x) = 6/(x - 2) + 4, we need to find the limit as x approaches infinity. As x gets very large, the fraction 6/(x - 2) approaches 0, leaving us with 0 + 4 = 4. Therefore, the horizontal asymptote is y = 4. Students might incorrectly choose y = 2 (choice B) by looking at the denominator's constant, or y = 0 (choice D) by only considering the fraction part. When a rational function is shifted vertically by adding a constant, that constant becomes the horizontal asymptote if the original function approaches zero.
For $q(x)=\dfrac{2x^4-x}{-x^4+6}$ , what is the horizontal asymptote as $x\to\infty$ ?
Answer: $y=-2$ This problem involves finding the horizontal asymptote of a rational function with equal degrees in numerator and denominator. For q(x) = (2x⁴ - x)/(-x⁴ + 6), both have degree 4, so we divide the leading coefficients: 2/(-1) = -2. Therefore, the horizontal asymptote is y = -2. Students might mistakenly choose y = 2 (choice B) by forgetting the negative sign in the denominator's leading coefficient, or y = 0 (choice D) thinking the degrees are different. When both polynomials have the same degree, always divide the coefficients of the highest degree terms, paying careful attention to signs.
For $s(x)=\dfrac{\sqrt{x^2+4x}}{x}$ , what is $\lim_{x\to\infty} s(x)$ ?
Answer: $1$ 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.
Let $u(x)=\dfrac{8}{5+e^x}$ . What is $\lim_{x\to\infty} u(x)$ ?
Answer: $0$ This problem involves finding limits at infinity to determine horizontal asymptotes for exponential functions. As x approaches infinity, $e^x$ grows without bound, so the denominator 5 + $e^x$ approaches infinity. Thus, u(x) = 8 / (5 + $e^x$ ) approaches 8 / infinity = 0. This means the horizontal asymptote is y = 0. A tempting distractor like 8/5 might occur if someone ignores the $e^x$ term and just divides by 5. A transferable strategy for horizontal asymptotes involving exponentials is to identify the dominant term in the denominator as x grows large, leading to a limit of zero when the numerator is constant.
For $s(x)=\dfrac{\sqrt{x^2+9}}{x}$ , what is $\lim_{x\to\infty} s(x)$ ?
Answer: $1$ This problem tests limits at infinity for $s(x)=\frac{\sqrt{x^2+9}}{x}$ . To evaluate $\lim_{x\to\infty} s(x)$ , we need to analyze the behavior of the square root term. For large positive $x$ , we can factor out $x^2$ from under the radical: $\sqrt{x^2+9}=\sqrt{x^2(1+\frac{9}{x^2})}=|x|\sqrt{1+\frac{9}{x^2}}=x\sqrt{1+\frac{9}{x^2}}$ (since $x>0$ as $x\to\infty$ ). Therefore, $s(x)=\frac{x\sqrt{1+\frac{9}{x^2}}}{x}=\sqrt{1+\frac{9}{x^2}}$ . As $x\to\infty$ , $\frac{9}{x^2}\to 0$ , so $s(x)\to\sqrt{1+0}=1$ . Choice D ( $9$ ) might tempt students who incorrectly think the $+9$ dominates. When square roots contain polynomial expressions, factor out the highest power of $x$ to find the limit at infinity.
For $v(x)=\dfrac{5x^3+2x}{x^3-100}$ , find the horizontal asymptote as $x\to\infty$ .
Answer: $y=5$ This problem involves finding limits at infinity to determine horizontal asymptotes. For v(x) = (5x³ + 2x)/(x³ - 100), the degrees are both 3, so the horizontal asymptote as x approaches infinity is y = 5/1 = 5. Odd degree but signs align for the same limit from both sides. Dominant terms prevail. A tempting distractor is y = 1/5, swapping the ratio. To find horizontal asymptotes of rational functions, compare the degrees of the numerator and denominator and, if equal, divide the leading coefficients.
For $h(x)=\dfrac{7-2x^3}{5x^3+x}$ , what is the horizontal asymptote as $x\to-\infty$ ?
Answer: $y=-\dfrac{2}{5}$ 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.
Let $r(x)=\dfrac{\ln(x)}{x}$ . Which value equals $\lim_{x\to\infty} r(x)$ ?
Answer: $0$ 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.
If $r(x)=\dfrac{-6x+9}{3x-1}$ , what is the horizontal asymptote of $r$ as $x\to\infty$ ?
Answer: $y=-2$ This question requires finding the horizontal asymptote of r(x) = (-6x + 9)/(3x - 1) as x approaches infinity, which tests limits at infinity. Both numerator and denominator have degree 1, so we factor out x from each: x(-6 + 9/x)/x(3 - 1/x) = (-6 + 9/x)/(3 - 1/x). As x approaches infinity, the terms 9/x and 1/x approach 0, leaving us with -6/3 = -2, so the horizontal asymptote is y = -2. Students might incorrectly choose y = 2 by dropping the negative sign or y = 0 by misunderstanding degree comparison. When degrees are equal in rational functions, divide the leading coefficients to find the horizontal asymptote.
Let $p(x)=\dfrac{2x^3-x}{x^3+4x^2+1}$ . What is the horizontal asymptote as $x\to\infty$ ?
Answer: $y=2$ This question tests finding horizontal asymptotes through limits at infinity for the function p(x) = (2x³ - x)/(x³ + 4x² + 1). Both numerator and denominator have degree 3, so we divide all terms by x³ to get (2 - 1/x²)/(1 + 4/x + 1/x³). As x approaches infinity, all terms with x in the denominator vanish, leaving us with 2/1 = 2, so the horizontal asymptote is y = 2. Students might mistakenly choose y = 1/2 by inverting the coefficient ratio or y = 0 by thinking the lower-degree terms dominate. For rational functions where numerator and denominator have equal highest degrees, the horizontal asymptote equals the ratio of leading coefficients.
Let $g(x)=\dfrac{5-2x}{x+4}$ . Which value equals $\lim_{x\to\infty} g(x)$ ?
Answer: $-2$ This problem involves finding limits at infinity to determine horizontal asymptotes for rational functions. To evaluate the limit of g(x) as x approaches infinity, divide both the numerator and denominator by x, resulting in (5/x - 2)/(1 + 4/x). As x approaches infinity, 5/x and 4/x approach zero, so the limit is -2/1 = -2. This value represents the horizontal asymptote y = -2. A tempting distractor like 0 might occur if someone incorrectly assumes the limit is zero because the denominator grows without considering the leading terms. 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.
Let $t(x)=\dfrac{x^2-4x+1}{2x^2+7}$ . What is $\lim_{x\to -\infty} t(x)$ ?
Answer: $\dfrac{1}{2}$ 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.
For $F(x)=\dfrac{3x^2+7}{6x^2-5}$ , what is the horizontal asymptote of $F$ ?
Answer: $y=\dfrac{1}{2}$ 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.
Let $g(x)=\dfrac{5x-4}{x^2+1}$ . What is the horizontal asymptote of $g$ as $x\to\infty$ ?
Answer: $y=0$ This question requires finding the horizontal asymptote of g(x) = (5x - 4)/(x² + 1) as x approaches infinity. The numerator has degree 1 while the denominator has degree 2, so the denominator grows much faster than the numerator. When the denominator's degree exceeds the numerator's degree, the horizontal asymptote is always y = 0. As x becomes very large, the function behaves like 5x/x² = 5/x, which approaches 0. Choice A (y = 5) incorrectly treats this as if the degrees were equal. For rational functions, when the denominator's degree is higher, the horizontal asymptote is y = 0.

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