AP Calculus AB: Continuity (Drill 1)

Question 1. Which of the following conditions is NOT required for a function \( f \) to be continuous at \( x = a \)?

• A) \( f(a) \) is defined
• B) \( \lim_{x \to a} f(x) \) exists
• C) \( f'(a) \) exists ✓
• D) \( \lim_{x \to a} f(x) = f(a) \)

Explanation: Choice C is correct. The three conditions for continuity at \( x = a \) are: (1) \( f(a) \) is defined, (2) \( \lim_{x \to a} f(x) \) exists, and (3) \( \lim_{x \to a} f(x) = f(a) \). Differentiability, the existence of \( f'(a) \), is NOT required. A function can be continuous without being differentiable (for example, \( f(x) = |x| \) is continuous but not differentiable at \( x = 0 \)). Choice A is incorrect because \( f(a) \) must be defined for continuity. Choice B is incorrect because the limit must exist. Choice D is incorrect because the limit must equal the function value.

Question 2. The function \( f(x) = \dfrac{x^2 - 9}{x - 3} \) has what type of discontinuity at \( x = 3 \)?

• A) Jump discontinuity
• B) Infinite discontinuity
• C) Removable discontinuity ✓
• D) The function is continuous at \( x = 3 \)

Explanation: Choice C is correct. Factoring gives \( \dfrac{(x-3)(x+3)}{x-3} = x + 3 \) for \( x \neq 3 \). The limit as \( x \to 3 \) equals 6, but \( f(3) \) is undefined (division by zero). Since the limit exists but the function is not defined at the point, this is a removable discontinuity; it could be repaired by defining \( f(3) = 6 \). Choice A doesn't work because a jump discontinuity requires the one-sided limits to exist but be unequal. Choice B falls short because an infinite discontinuity requires the limit to be \( \pm\infty \), which is not the case here. Choice D is incorrect because \( f(3) \) is undefined.

Question 3. Let \( f \) be defined by \( f(x) = x^2 + 1 \) for \( x < 2 \) and \( f(x) = kx - 1 \) for \( x \geq 2 \). For what value of \( k \) is \( f \) continuous at \( x = 2 \)?

• A) \( k = 1 \)
• B) \( k = 2 \)
• C) \( k = 3 \) ✓
• D) \( k = 6 \)

Explanation: Choice C is correct. The left-hand limit at \( x = 2 \) is \( \lim_{x \to 2^-}(x^2 + 1) = 5 \). The function value from the right branch is \( f(2) = 2k - 1 \). Setting \( 2k - 1 = 5 \) gives \( 2k = 6 \), so \( k = 3 \). Choice A doesn't fit: \( k = 1 \) gives \( f(2) = 1 \), not 5. Choice B is wrong because \( k = 2 \) gives \( f(2) = 3 \), not 5. Choice D doesn't work because \( k = 6 \) results from forgetting that \( x = 2 \) is substituted into \( kx - 1 \) and instead solving \( k - 1 = 5 \) as if the coefficient of \( k \) were 1.

Question 4. A student claims: “If \( \lim_{x \to 4} f(x) = 7 \), then \( f \) must be continuous at \( x = 4 \).” Which of the following best explains why this claim is incorrect?

• A) The limit of \( f \) at \( x = 4 \) could be different depending on the direction of approach.
• B) The existence of \( \lim_{x \to 4} f(x) \) does not guarantee that \( f(4) \) is defined or equals 7. ✓
• C) Continuity at a point requires the limit to be zero.
• D) The limit existing at \( x = 4 \) means \( f \) must be differentiable there, not just continuous for the function in question.

Explanation: Choice B is correct. Continuity at \( x = 4 \) requires three conditions: the limit must exist, the function must be defined at 4, and the limit must equal the function value. The student’s claim only verifies that the limit exists and equals 7, but \( f(4) \) could be undefined (removable discontinuity) or defined as a value other than 7, either of which would make \( f \) discontinuous at \( x = 4 \). Choice A falls short because if the two-sided limit equals 7, both one-sided limits already agree. Choice C misses the mark: there is no requirement that the limit equal zero. Choice D is off because differentiability is a separate, stronger condition than continuity, the implication goes the other way (differentiability implies continuity, not vice versa).

Question 5. Let \( f \) be defined by \( f(x) = \dfrac{x^2 - 4}{x - 2} \) for \( x \neq 2 \) and \( f(2) = 5 \). Which of the following statements about \( f \) is true?

• A) \( f \) is continuous at \( x = 2 \) because \( f(2) \) is defined.
• B) \( f \) is continuous at \( x = 2 \) because \( \lim_{x \to 2} f(x) \) exists.
• C) \( f \) is not continuous at \( x = 2 \) because \( f(2) \neq \lim_{x \to 2} f(x) \). ✓
• D) \( f \) is not continuous at \( x = 2 \) because \( \lim_{x \to 2} f(x) \) does not exist.

Explanation: Choice C is correct. For \( x \neq 2 \): \( \dfrac{x^2 - 4}{x - 2} = \dfrac{(x-2)(x+2)}{x-2} = x + 2 \), so \( \lim_{x \to 2} f(x) = 4 \). However, \( f(2) = 5 \). Since \( 4 \neq 5 \), the third continuity condition fails; this is a removable discontinuity that has not been correctly repaired (the function is defined at the point, but at the wrong value). Choice A is wrong because being defined alone is not sufficient for continuity. Choice B doesn't work because the limit existing alone is also not sufficient. Choice D falls short because the limit does exist; it equals 4.