Answer: $-3$

Handle the square root carefully: For $x < 0$, $\sqrt{x^2} = |x| = -x$. Factor inside the radical: $\sqrt{9x^2 + 1} = |x|\sqrt{9 + 1/x^2} = -x\sqrt{9 + 1/x^2}$ when $x < 0$. Compute the limit: $$\frac{\sqrt{9x^2+1}}{x} = \frac{-x\sqrt{9+1/x^2}}{x} = -\sqrt{9+1/x^2}$$ As $x \to -\infty$, $1/x^2 \to 0$, so the limit is $-\sqrt{9} = -3$. Why distractors fail: Option A ($3$) neglects the sign; this would be the answer for $x \to +\infty$. Option C ($0$) reflects a degree-comparison error. Option D is wrong because the limit exists and equals $-3$.

Answer: $-2$

Identify dominant terms: The numerator is dominated by $2x^3$. The denominator is $\sqrt{x^6 + 3} \approx \sqrt{x^6} = |x^3|$. Handle the absolute value: For $x \to -\infty$, $x^3 < 0$, so $|x^3| = -x^3$. Thus $\frac{2x^3}{|x^3|} = \frac{2x^3}{-x^3} = -2$... but let's be more careful. Divide numerator and denominator by $|x^3| = -x^3$: $$\frac{2x^3/(-x^3) + x/(-x^3)}{\sqrt{x^6+3}/(-x^3)}$$ Numerator: $-2 + (-1/x^2) \to -2$. Denominator: $\frac{\sqrt{x^6+3}}{-x^3}$. Since $x < 0$, $x^3 < 0$, so $-x^3 > 0$ and $\frac{\sqrt{x^6+3}}{-x^3} = \frac{\sqrt{x^6+3}}{|x^3|} = \sqrt{1 + 3/x^6} \to 1$. So the limit is $\frac{-2}{1} = -2$. Corrected answer: The correct limit is $-2$, not $2$. Let's verify once more: for $x < 0$, $\sqrt{x^6} = |x|^3 = (-x)^3 = -x^3$. So $\frac{2x^3}{\sqrt{x^6}} = \frac{2x^3}{-x^3} = -2$. The limit is $-2$. Why distractors fail: Option A ($2$) neglects the sign change from the absolute value when $x < 0$. Option C ($0$) incorrectly assumes the denominator grows faster. Option D is wrong because the limit exists.

Answer: $2$

Rationalize the expression: Multiply and divide by the conjugate: $$\frac{(\sqrt{x^2+4x} - x)(\sqrt{x^2+4x} + x)}{\sqrt{x^2+4x} + x} = \frac{x^2+4x - x^2}{\sqrt{x^2+4x}+x} = \frac{4x}{\sqrt{x^2+4x}+x}$$ Simplify and evaluate: Divide numerator and denominator by $x$ (positive for $x \to \infty$): $$\frac{4}{\sqrt{1+4/x}+1}$$ As $x \to \infty$, this becomes $\frac{4}{1+1} = 2$. Why distractors fail: Option A ($0$) is a common error from assuming the difference of two large quantities is zero. Option B ($4$) incorrectly uses the coefficient without completing the algebra. Option D ($\infty$) reflects an $\infty - \infty$ misconception.

Answer: $-2$

Apply dominant term analysis: Both numerator and denominator have degree 2. The limit equals the ratio of leading coefficients: $\frac{4}{-2} = -2$. Why distractors fail: Option A ($2$) ignores the negative sign in the denominator's leading coefficient. Option C ($0$) would apply if the numerator degree were less. Option D ($\infty$) would apply if the numerator degree were greater.

Answer: $+\infty$

Analyze signs near $x = 2^+$: As $x \to 2^+$: the numerator $x + 1 \to 3 > 0$; the factor $(x - 2) \to 0^+$ (positive, approaching from the right); the factor $(x + 3) \to 5 > 0$. Determine the limit: All factors are positive, and the denominator approaches $0^+$, so the expression approaches $\frac{3}{0^+} = +\infty$. Why distractors fail: Option B ($-\infty$) would occur if the sign combination were negative. Option C substitutes $x = 2$ directly without recognizing the infinite limit. Option D ($0$) has no justification here.

Answer: The function simplifies to $f(x) = x - 2$ for all $x$, so there are no vertical asymptotes.

Factor the numerator: Recognize $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$ using the difference of cubes formula $a^3 - b^3 = (a-b)(a^2+ab+b^2)$. Cancel common factors: $$f(x) = \frac{(x-2)(x^2+2x+4)}{x^2+2x+4} = x - 2$$ for all $x$ where the denominator is nonzero. The discriminant of $x^2+2x+4$ is $4 - 16 = -12 < 0$, so the denominator has no real zeros. The function equals $x - 2$ everywhere. Why distractors fail: Option A incorrectly claims the denominator has real zeros. Option C is partially correct about no real zeros but misses the key fact that the function simplifies completely. Option D invents a vertical asymptote at $x = 2$, which is just a point on the line $y = x - 2$.

Answer: $y = 3$

Compare degrees of numerator and denominator: Both the numerator $3x^2 + 1$ and the denominator $x^2 - 4$ have degree 2. When degrees are equal, the horizontal asymptote is the ratio of the leading coefficients. Calculate the ratio: The leading coefficient of the numerator is 3 and of the denominator is 1, so the horizontal asymptote is $y = \frac{3}{1} = 3$. Why distractors fail: Option A ($y = 0$) applies when the numerator degree is less than the denominator degree. Option C confuses the constant in the denominator with the asymptote. Option D is incorrect because equal-degree rational functions always have a horizontal asymptote.

Answer: $0$

Compare degrees: The numerator has degree 1 and the denominator has degree 2. When the degree of the denominator exceeds the degree of the numerator, the limit at infinity is 0. Why Option B is correct: Dividing numerator and denominator by $x^2$ gives $\frac{5/x}{2 + 7/x^2}$. As $x \to \infty$, the numerator approaches 0 and the denominator approaches 2, so the limit is 0. Why distractors fail: Option A ($5/2$) would be the answer if both had degree 1. Option C ($\infty$) would apply if the numerator had higher degree. Option D incorrectly uses the constants 5 and 7.

Answer: Yes, the approach and answer are both correct.

Verify the $\infty/\infty$ form: As $x \to \infty$, both numerator and denominator approach infinity, creating an $\frac{\infty}{\infty}$ indeterminate form. L'Hôpital's Rule applies. Check the differentiation: Differentiating: numerator becomes $6x$, denominator becomes $4x$. The ratio $\frac{6x}{4x} = \frac{3}{2}$ for all $x \neq 0$, so $\lim_{x \to \infty} \frac{6x}{4x} = \frac{3}{2}$. Confirm the answer matches the standard method: By dominant term analysis, the ratio of leading coefficients is $\frac{3}{2}$, confirming the L'Hôpital's Rule result. Both methods are valid. Why distractors fail: Option A is wrong; L'Hôpital's Rule applies to $\infty/\infty$ forms. Option C presents $6/4$ and $3/2$ as different numbers, but they're equal. Option D is wrong because after one application, the limit is already determinate.

Answer: The horizontal asymptote is the ratio of the leading coefficients.

Equal-degree rule: When the degrees of the numerator and denominator are equal, the horizontal asymptote is $y = \frac{a_n}{b_n}$ where $a_n$ and $b_n$ are the leading coefficients. Why distractors fail: Option A is wrong because equal degrees do yield a horizontal asymptote. Option B ($y = 0$) applies only when the numerator degree is strictly less. Option D (oblique asymptote) applies when the numerator degree exceeds the denominator degree by exactly one.

Answer: $f(x) = \frac{5x^2}{x^2 + 1}$

Check requirements for each option: We need: (1) no vertical asymptotes (denominator never zero for real $x$), (2) HA at $y = 5$, (3) defined for all real $x$. Verify Option B: $f(x) = \frac{5x^2}{x^2+1}$: The denominator $x^2 + 1 > 0$ for all real $x$ ✓. Equal degrees with ratio $5/1 = 5$, so HA is $y = 5$ ✓. Defined everywhere ✓. Why distractors fail: Option A has a vertical asymptote at $x = 1$. Option C has an undefined point at $x = 0$ (and HA at $y = 0$, not $y = 5$). Option D is a linear function with no horizontal asymptote.

Answer: $f(x) \to \infty$ as $x \to \pm\infty$, with end behavior resembling $|x|$

Analyze for large positive $x$: For $x > 0$: $|x| = x$, so $f(x) = \frac{x^2+1}{x+1}$. The numerator has degree 2 and denominator degree 1, so $f(x) \to \infty$. More precisely, $f(x) \approx x - 1$ by long division. Analyze for large negative $x$: For $x < 0$: $|x| = -x$, so $f(x) = \frac{x^2+1}{-x+1}$. Again the numerator has degree 2 and denominator degree 1, so $f(x) \to \infty$ as $x \to -\infty$ as well. The function grows like $|x|$ in both directions. Why distractors fail: Option A is wrong because the numerator degree exceeds the denominator degree, so there's no horizontal asymptote. Option B is wrong because $f(x)$ is always positive (both numerator and denominator are positive). Option D is wrong for the same reason as A.

Answer: $\infty$

Compare growth rates: Exponential functions like $e^x$ grow faster than any polynomial $x^n$ as $x \to \infty$. Therefore $\frac{e^x}{x^3} \to \infty$. Why distractors fail: Option A ($0$) reverses the growth rate relationship. Option B ($1$) has no basis for this function. Option D is wrong because the limit does exist (it is $+\infty$).

Answer: $0$

Growth rate comparison: The logarithmic function $\ln x$ grows much slower than the linear function $x$ as $x \to \infty$. Therefore the ratio $\frac{\ln x}{x} \to 0$. Why distractors fail: Option A ($1$) has no basis. Option B ($\infty$) reverses the growth rate dominance. Option D ($-\infty$) is impossible since both $\ln x$ and $x$ are positive for large $x$.

Answer: A removable discontinuity (hole)

Factor and simplify: The numerator factors as $(x-1)(x+1)$, so $h(x) = \frac{(x-1)(x+1)}{x-1} = x+1$ for $x \neq 1$. The common factor cancels. Classify the discontinuity: Since the limit $\lim_{x \to 1} h(x) = 2$ exists and is finite, but $h(1)$ is undefined, this is a removable discontinuity (hole), not a vertical asymptote. Why distractors fail: Option A is wrong because the factor causing the zero in the denominator cancels; the function does not blow up. Option C (jump discontinuity) requires different one-sided limits. Option D is wrong because $h(1)$ is undefined.

Answer: $x = 3$ only

Factor and simplify: The denominator factors as $(x-3)(x+3)$, and the numerator is $(x+3)$. The common factor $(x+3)$ cancels, leaving $\frac{1}{x-3}$ with a hole at $x = -3$. Identify the vertical asymptote: After cancellation, the only factor remaining in the denominator is $(x - 3)$, so the vertical asymptote is $x = 3$ only. Why distractors fail: Option A includes $x = -3$, but this is a removable discontinuity (hole), not an asymptote. Option C incorrectly identifies the cancelled factor as the asymptote. Option D confuses the constant 9 in the denominator with the asymptote location.

Answer: The claim is incorrect; $x = 2$ is a removable discontinuity, not a vertical asymptote.

Factor both expressions: Numerator: $x^2 - 4 = (x-2)(x+2)$. Denominator: $x^2 - x - 6 = (x-3)(x+2)$. Identify common factors: Wait — let's recheck. The denominator factors as $(x-3)(x+2)$. The numerator factors as $(x-2)(x+2)$. The common factor is $(x+2)$, which cancels. After cancellation: $f(x) = \frac{x-2}{x-3}$ for $x \neq -2$. At $x = 2$, the simplified function gives $\frac{0}{-1} = 0$, which is finite. But $x = 2$ was not a zero of the original denominator either: $2^2 - 2 - 6 = -4 \neq 0$. So $x = 2$ is simply a zero of the function, not a discontinuity at all. Corrected analysis: Actually, $x = 2$ makes the denominator $4 - 2 - 6 = -4 \neq 0$. So the student's claim is wrong because $x = 2$ is not even a zero of the denominator — it's just an ordinary point where $f(2) = 0$. The vertical asymptote is at $x = 3$ and the removable discontinuity is at $x = -2$. Why the correct answer works: Option B is the best answer. The student confused $x = 2$ (a zero of the numerator) with a vertical asymptote. There is no vertical asymptote or any discontinuity at $x = 2$. Why distractors fail: Option A is wrong because the denominator is not zero at $x = 2$. Option C is wrong because there is a vertical asymptote at $x = 3$. Option D is wrong because both being zero does not confirm an asymptote — and in this case, only the numerator is zero at $x = 2$.

Answer: $f(x) = \frac{2x}{\sqrt{x^2 + 1}}$

Test Option B: For $x \to +\infty$: $\frac{2x}{\sqrt{x^2+1}} = \frac{2x}{|x|\sqrt{1+1/x^2}} = \frac{2x}{x \cdot 1} = 2$. For $x \to -\infty$: $|x| = -x$, so $\frac{2x}{-x} = -2$. Thus horizontal asymptotes are $y = 2$ and $y = -2$. Why distractors fail: Option A: both limits equal 0. Option C: both limits equal 1. Option D: both limits equal 0. Only Option B has different horizontal asymptotes because the square root introduces a sign-dependent behavior.

Answer: The expression simplifies to $\frac{5}{x^2 - 1}$, and the limit is $0$.

Combine into a single fraction: $$\frac{3x^2+2}{x^2-1} - 3 = \frac{3x^2+2 - 3(x^2-1)}{x^2-1} = \frac{3x^2 + 2 - 3x^2 + 3}{x^2-1} = \frac{5}{x^2-1}$$ Evaluate the limit: As $x \to \infty$, $\frac{5}{x^2-1} \to 0$. This confirms the horizontal asymptote is exactly $y = 3$ and the function approaches it from above. Why distractors fail: Option A gives the correct numerical limit but doesn't show the work. Option C is the best answer because it shows the simplification and the limit. Option B ($5$) is the numerator of the simplified expression, not the limit. Option D ($3$) is the horizontal asymptote of the original function, not the limit of the difference.

Answer: $y = 2x + 1$

Confirm oblique asymptote exists: The numerator has degree 2 and the denominator has degree 1. Since the difference is exactly 1, there is an oblique asymptote. Perform polynomial long division: $2x^2 + 3x - 5$ divided by $x + 1$: First term: $2x^2 \div x = 2x$. Multiply: $2x(x+1) = 2x^2 + 2x$. Subtract: $(3x - 5) - 2x = x - 5$. Second term: $x \div x = 1$. Multiply: $1(x+1) = x+1$. Subtract: $(x-5) - (x+1) = -6$. So $f(x) = 2x + 1 + \frac{-6}{x+1}$, and the oblique asymptote is $y = 2x + 1$. Why distractors fail: Options A, B, and D result from arithmetic errors during the long division process. Careful subtraction at each step is essential.

Answer: $f(x) = \frac{2x^2}{(x+1)(x-3)}$

Requirements analysis: We need: (1) denominator zeros at $x = -1$ and $x = 3$, so denominator is $(x+1)(x-3) = x^2 - 2x - 3$; (2) HA at $y = 2$, so numerator must have degree 2 with leading coefficient $2 \cdot 1 = 2$; (3) $f(0) = 0$. Check each option at $x = 0$: Option A: $f(0) = \frac{0}{(1)(-3)} = 0$ ✓. Option B: $f(0) = \frac{0}{-3} = 0$ ✓. Option C: $f(0) = \frac{0}{(1)(-3)} = 0$ ✓. Option D: $f(0) = \frac{0}{-3} = 0$ ✓. All pass through the origin, so we need further checks. Check that numerator doesn't cancel with denominator: Option A: $\frac{2x^2}{x^2-2x-3}$. At $x = -1$: numerator $= 2 \neq 0$, and at $x = 3$: numerator $= 18 \neq 0$ ✓. Option B: $2x^2+6x = 2x(x+3)$. At $x = -1$: $2(-1)(2) = -4 \neq 0$ ✓, at $x = 3$: $2(3)(6) = 36 \neq 0$ ✓. Option C: $2x(x+3)$, same as B. Option D: $2x^2-6x = 2x(x-3)$. At $x = 3$: numerator $= 0$! The factor $(x-3)$ cancels, removing the vertical asymptote at $x = 3$. Recheck — correct answer identification: Wait — Option D has $2x(x-3)$ in the numerator, which cancels $(x-3)$ in the denominator. This means $x = 3$ becomes a hole, not a VA. So Option D fails. Let's recheck Option A: $\frac{2x^2}{(x+1)(x-3)}$ — VAs at $x=-1,3$ ✓, HA $y=2$ ✓, $f(0)=0$ ✓. This appears to satisfy all conditions. However, we should verify all options more carefully. Option A satisfies all three conditions. Let me re-examine: Option A is $\frac{2x^2}{(x+1)(x-3)}$, which has leading coefficient ratio $2/1 = 2$ ✓, denominator zeros at $-1$ and $3$ with no cancellation ✓, and $f(0) = 0$ ✓. Final resolution: Upon careful analysis, Option A $\frac{2x^2}{(x+1)(x-3)}$ satisfies all conditions: VAs at $x = -1$ and $x = 3$, HA at $y = 2$, and $f(0) = 0$. Option D fails because the factor $(x-3)$ cancels. The correct answer is Option A.

Answer: The long-term carrying capacity of 500 individuals

Find the horizontal asymptote: $\lim_{t \to \infty} \frac{500t}{t+20} = \frac{500}{1} = 500$. The horizontal asymptote is $P = 500$. Interpret in context: As time increases, the population approaches but never exceeds 500. In ecological modeling, this represents the carrying capacity — the long-term sustainable population. Why distractors fail: Option A: $P(0) = 0$, which is the initial population, not the asymptote. Option B confuses the asymptote with a rate. Option D confuses a population level with a time value.

Answer: A line $x = a$ such that $\lim_{x \to a^+} f(x) = \pm\infty$ or $\lim_{x \to a^-} f(x) = \pm\infty$

Definition of a vertical asymptote: A vertical asymptote occurs at $x = a$ when at least one of the one-sided limits of $f(x)$ approaches $+\infty$ or $-\infty$ as $x$ approaches $a$. Why distractors fail: Option A describes a horizontal asymptote. Option C describes an oblique (slant) asymptote. Option D describes a zero of the function, not an asymptote — a function equals zero at its roots, but the function must blow up to $\pm\infty$ for a vertical asymptote.

Answer: The function's values get arbitrarily close to $L$ as $x$ increases or decreases without bound.

Meaning of horizontal asymptote: A horizontal asymptote $y = L$ means $\lim_{x \to \infty} f(x) = L$ or $\lim_{x \to -\infty} f(x) = L$. The function values approach $L$ but need not equal it. Why distractors fail: Option A is wrong because the function approaches but need not equal $L$. Option C is a common misconception — functions can cross horizontal asymptotes (e.g., $f(x) = \frac{\sin x}{x}$ crosses $y = 0$ infinitely often). Option D confuses asymptotic behavior with extreme values.

Answer: $\frac{-3x + 1}{x - 4}$

Use horizontal asymptote condition: For $\frac{ax+b}{cx+d}$, the horizontal asymptote is $y = a/c$. We need $a/c = -3$. Use vertical asymptote condition: The vertical asymptote is at $x = -d/c$. We need $-d/c = 4$, so $d = -4c$. With $c = 1$, this gives $d = -4$. Verify Option A: For $\frac{-3x+1}{x-4}$: horizontal asymptote is $y = -3/1 = -3$ ✓, and vertical asymptote is $x = 4$ ✓. Why distractors fail: Option B has horizontal asymptote $y = 3$ (wrong sign). Option C has vertical asymptote at $x = -4$. Option D has horizontal asymptote $y = 1$, not $-3$.

Answer: No, because the function has an oblique asymptote $y = 2x$, found by polynomial long division.

Degree comparison: The numerator has degree 3 and the denominator has degree 2. Since the numerator degree exceeds the denominator degree by exactly 1, there is an oblique (slant) asymptote. Perform polynomial long division: Dividing $2x^3 + x$ by $x^2 + 1$: $2x^3 \div x^2 = 2x$. Then $2x(x^2+1) = 2x^3 + 2x$. Subtracting: $(2x^3 + x) - (2x^3 + 2x) = -x$. The remainder is $\frac{-x}{x^2+1} \to 0$ as $x \to \infty$, so the oblique asymptote is $y = 2x$. Why distractors fail: Option A incorrectly claims no asymptote exists. Option B incorrectly identifies a horizontal asymptote (which doesn't exist when the numerator degree exceeds the denominator). Option C gives an incorrect oblique asymptote $y = 2x - 1$.