AP Calculus AB: Limits Involving Infinity and Special Cases (Drill 1)

Question 1. What is \( \lim_{x \to \infty} \dfrac{4x^2 - 3x + 1}{2x^2 + 5} \)?

• A) \( 0 \)
• B) \( 2 \) ✓
• C) \( 4 \)
• D) The limit does not exist.

Explanation: Choice B is correct. When the numerator and denominator have equal degree, the limit at infinity equals the ratio of leading coefficients: \( \dfrac{4}{2} = 2 \). Choice A is wrong because a limit of 0 results when the numerator’s degree is less than the denominator’s, the student has the rule backwards. Choice C doesn't work because reporting only the leading coefficient of the numerator (4) without dividing by the denominator’s leading coefficient (2). Choice D falls short because the limit exists and equals 2; a horizontal asymptote exists whenever the degrees are equal.

Question 2. What is \( \lim_{x \to 3^+} \dfrac{5}{x - 3} \)?

• A) \( -\infty \)
• B) \( 0 \)
• C) \( 5 \)
• D) \( +\infty \) ✓

Explanation: Choice D is correct. As \( x \to 3^+ \), the quantity \( (x-3) \to 0^+ \) (small positive values). Therefore \( \dfrac{5}{x-3} \to +\infty \). Choice A misses the mark: \( -\infty \) results from the left-hand limit \( x \to 3^- \), where \( (x-3) \to 0^- \); this confuses the two one-sided limits. Choice B is off because treating the numerator value at \( x = 3 \) as the limit result. Choice C doesn't fit: substituting \( x = 3 \) into the numerator only and ignoring the denominator behavior.

Question 3. What is \( \lim_{x \to \infty} \dfrac{6x^3 + 2x}{x^4 - 1} \)?

• A) \( 0 \) ✓
• B) \( 6 \)
• C) \( \infty \)
• D) \( -\infty \)

Explanation: Choice A is correct. The numerator’s degree (3) is less than the denominator’s degree (4), so the denominator grows faster and the limit equals 0. Choice B is off because applying the equal-degree rule (ratio of leading coefficients \( = \dfrac{6}{1} \)) when the degrees are not equal. Choice C doesn't fit: reversing the degree comparison, \( \infty \) would result if the numerator’s degree exceeded the denominator’s. Choice D is wrong because a student who confuses the sign of the result based on the direction of growth; since all leading terms are positive, the expression approaches 0 from the positive side, never \( -\infty \).

Question 4. The function \( f \) is defined as \( f(x) = \dfrac{|x - 2|}{x - 2} \). Which of the following correctly describes \( \lim_{x \to 2} f(x) \)?

• A) The limit equals 1.
• B) The limit equals −1.
• C) The limit equals 0.
• D) The limit does not exist. ✓

Explanation: Choice D is correct. For \( x > 2 \): \( |x-2| = x-2 \), so \( f(x) = \dfrac{x-2}{x-2} = 1 \). For \( x < 2 \): \( |x-2| = -(x-2) \), so \( f(x) = \dfrac{-(x-2)}{x-2} = -1 \). The right-hand limit is 1 and the left-hand limit is −1. Since the one-sided limits are unequal, the two-sided limit does not exist. Choice A misses the mark: this is only the right-hand limit \( \lim_{x \to 2^+} f(x) = 1 \); the student ignores left-hand behavior. Choice B is off because this is only the left-hand limit \( \lim_{x \to 2^-} f(x) = -1 \); the student ignores right-hand behavior. Choice C doesn't fit: averaging the one-sided limits \( \dfrac{1 + (-1)}{2} = 0 \) is not a valid limit technique.

Question 5. A rational function \( g(x) = \dfrac{3x^2 + 7}{ax^2 + bx + 1} \) has a horizontal asymptote of \( y = \dfrac{3}{5} \). What must be true?

• A) \( a = 3 \) and \( b = 5 \)
• B) \( b = 5 \) and \( a \) can be any nonzero value
• C) \( a = 5 \) ✓
• D) \( a = 5 \) and \( b = 0 \)

Explanation: Choice C is correct. Since numerator and denominator have equal degree, the horizontal asymptote equals the ratio of leading coefficients: \( \dfrac{3}{a} = \dfrac{3}{5} \), so \( a = 5 \). The value of \( b \) has no effect on the horizontal asymptote, only leading coefficients determine end behavior. Choice A doesn't work because \( b \) is irrelevant to the horizontal asymptote; including a condition on \( b \) reflects a misconception about which terms control end behavior. Choice B falls short because \( a \) is uniquely determined as 5; it cannot be any nonzero value. Choice D misses the mark: \( a = 5 \) is necessary, but \( b \) is not required to equal 0, for example, \( g(x) = \dfrac{3x^2+7}{5x^2+100x+1} \) still has horizontal asymptote \( y = \dfrac{3}{5} \).