AP Calculus AB 1.10 Exploring Types of 3 Discontinuities - MCQs - Exam Style Questions
No-Calc Question
The function \(f\) is defined by \( f(x)=\frac{|\sin x|}{x^{3}-1}. \) At how many values of \(x\) does \(f\) have a discontinuity?
(A) Zero
(B) One
(C) Three
(D) Infinite
(B) One
(C) Three
(D) Infinite
▶️ Answer/Explanation
\(|\sin x|\) is continuous for all real \(x\).
Discontinuities occur only where the denominator is \(0\): \(x^{3}-1=0 \Rightarrow x=1\).
Thus \(f\) is discontinuous at exactly one \(x\)-value.
✅ Answer: (B) One
Discontinuities occur only where the denominator is \(0\): \(x^{3}-1=0 \Rightarrow x=1\).
Thus \(f\) is discontinuous at exactly one \(x\)-value.
✅ Answer: (B) One
▶️ Answer/Explanation
Solution
Correct Answer: C
Analysis of the function at x = 2:
Left-hand limit (x→2⁻): \(\lim_{x \to 2^-}\frac{x-2}{2|x-2|} = \lim_{x \to 2^-}\frac{x-2}{-2(x-2)} = -\frac{1}{2}\)
Right-hand limit (x→2⁺): \(\lim_{x \to 2^+}\frac{x-2}{2|x-2|} = \lim_{x \to 2^+}\frac{x-2}{2(x-2)} = \frac{1}{2}\)
Conclusion:
1. The left and right limits are not equal (-½ ≠ ½), so the overall limit does not exist
2. The function has a jump discontinuity at x = 2 (finite jump between -½ and ½)
3. There is no vertical asymptote (the function values remain finite)