AP Calculus AB: Squeeze Theorem, IVT, and Mixed Limit Skills (Drill 1)

Question 1. Using the Squeeze Theorem, what is \( \lim_{x \to 0} x^2 \sin\!\left(\dfrac{1}{x}\right) \)?

• A) The limit does not exist.
• B) \( 1 \)
• C) \( 0 \) ✓
• D) \( -1 \)

Explanation: Choice C is correct. Since \( -1 \leq \sin\!\left(\dfrac{1}{x}\right) \leq 1 \) for all \( x \neq 0 \), multiplying through by \( x^2 \geq 0 \) gives \( -x^2 \leq x^2\sin\!\left(\dfrac{1}{x}\right) \leq x^2 \). As \( x \to 0 \), both \( -x^2 \to 0 \) and \( x^2 \to 0 \), so by the Squeeze Theorem the limit equals 0. Choice A is off because students who notice that \( \sin(1/x) \) oscillates near 0 may conclude the product has no limit, but the \( x^2 \) factor forces the entire expression to zero regardless of the oscillation. Choice B doesn't fit: this confuses the bounded range of \( \sin(1/x) \) with the limit of the full product. Choice D is wrong because while the expression can take negative values, the limit from both sides is 0.

Question 2. The function \( f \) is continuous on \( [1, 5] \) with \( f(1) = -3 \) and \( f(5) = 8 \). Which of the following is guaranteed by the Intermediate Value Theorem?

• A) There exists \( c \in (1, 5) \) such that \( f(c) = 0 \). ✓
• B) \( f \) achieves a maximum value on \( [1, 5] \).
• C) \( f \) is differentiable on \( (1, 5) \).
• D) There is exactly one value \( c \) in \( (1, 5) \) such that \( f(c) = 0 \).

Explanation: Choice A is correct. The IVT states that if \( f \) is continuous on \( [a, b] \) and \( N \) is any value strictly between \( f(a) \) and \( f(b) \), then there exists at least one \( c \in (a, b) \) with \( f(c) = N \). Since \( f(1) = -3 < 0 < 8 = f(5) \), the value 0 lies between the two output values, so the IVT guarantees at least one root in \( (1, 5) \). Choice B is wrong because that is the conclusion of the Extreme Value Theorem, not the IVT. Choice C doesn't work because continuity does not imply differentiability. Choice D falls short because the IVT guarantees the existence of at least one such value; it says nothing about uniqueness. There could be one root, three roots, or infinitely many.

Question 3. Let \( f \) be defined by \( f(x) = 2x + 1 \) for \( x \leq 1 \) and \( f(x) = x^2 + bx \) for \( x > 1 \). For what value of \( b \) is \( f \) continuous at \( x = 1 \)?

• A) \( b = -1 \)
• B) \( b = 1 \)
• C) \( b = 2 \) ✓
• D) \( b = 3 \)

Explanation: Choice C is correct. The left-hand limit (and function value) at \( x = 1 \) from the first branch is \( 2(1) + 1 = 3 \). For continuity, the right-hand limit must also equal 3: \( \lim_{x \to 1^+}(x^2 + bx) = 1 + b \). Setting \( 1 + b = 3 \) gives \( b = 2 \). Choice A is off because \( b = -1 \) gives a right-hand limit of \( 1 + (-1) = 0 \), not 3. Choice B doesn't fit: \( b = 1 \) gives a right-hand limit of \( 1 + 1 = 2 \), not 3. Choice D is wrong because \( b = 3 \) gives a right-hand limit of \( 1 + 3 = 4 \), not 3; this error arises from dropping the \( x^2 \) term and treating the right branch as simply \( bx \), then solving \( b(1) = 3 \).

Question 4. A student wants to use the Intermediate Value Theorem to show that \( f(x) = x^3 - 4x + 1 \) has a root on the interval \( [0, 2] \). Which of the following correctly justifies this conclusion?

• A) \( f(0) = 1 > 0 \) and \( f(2) = 1 > 0 \), so the function must cross zero somewhere in between.
• B) \( f(0) = 1 \) and \( f(2) = 1 \), and since \( f \) is a polynomial it must equal zero somewhere between 0 and 2 based on the given information.
• C) \( f(0) = 1 > 0 \) and \( f(1) = -2 < 0 \), so since \( f \) is continuous, there is a root in \( (0, 1) \subset [0, 2] \). ✓
• D) \( f \) is a polynomial, so it is guaranteed to have at least one real root on any interval.

Explanation: Choice C is correct. \( f(0) = 0 - 0 + 1 = 1 > 0 \) and \( f(1) = 1 - 4 + 1 = -2 < 0 \). Since \( f \) is a polynomial (continuous everywhere), the IVT guarantees that every value between \( f(0) = 1 \) and \( f(1) = -2 \) is attained on \( (0, 1) \). Because \( 0 \) lies between \( -2 \) and \( 1 \), there is a root in \( (0, 1) \subset [0, 2] \). Choice A doesn't fit: \( f(2) = 8 - 8 + 1 = 1 \), so both endpoints give the same positive value. To guarantee that 0 is attained using endpoint data alone, the value 0 must lie strictly between the two output values, which it does not here since both outputs equal 1. Choice B is incorrect for the same reason; equal-sign endpoint values do not guarantee a root between them. Choice D is wrong because polynomials are not guaranteed to have roots on every interval (for example, \( f(x) = x^2 + 1 \) has no real roots on any interval).

Question 5. Suppose \( f \) is continuous on \( [-2, 6] \) with \( f(-2) = 4 \) and \( f(6) = 2 \). Let \( g(x) = f(x) - x \). Which of the following must be true?

(A) There exists \( c \in (-2, 6) \) such that \( f(c) = 3 \).
(B) There exists \( c \in (-2, 6) \) such that \( f(c) = c \).

• A) Statement (A) only
• B) Statement (B) only
• C) There exists \( c \in (-2, 6) \) such that \( f(c) = 0 \).
• D) Both statements (A) and (B) must be true. ✓

Explanation: Choice D is correct, both statements follow from the IVT. For statement (A): \( f \) is continuous on \( [-2, 6] \) with \( f(6) = 2 < 3 0 \) and \( g(6) = f(6) - 6 = 2 - 6 = -4 < 0 \). Since \( g \) changes sign, the IVT guarantees a \( c \in (-2, 6) \) where \( g(c) = 0 \), meaning \( f(c) = c \), a fixed point. Choice A is incorrect because statement (B) is also guaranteed. Choice B is incorrect because statement (A) is also guaranteed. Choice C is incorrect: the IVT guarantees that \( f \) attains every value in \( [2, 4] \) (between the two output values), and since \( 0 < 2 \), zero lies outside that range and is not guaranteed.