Exploring Types of Discontinuities - AP Calculus BC

Let $p(x)=\sin\!\left(\frac{1}{x}\right)$ for $x\ne 0$ and $p(0)=0$ . What type of discontinuity occurs at $x=0$ ?
Answer: Oscillating discontinuity This question involves classifying discontinuities in oscillating functions. For $p(x)=\sin\left(\frac{1}{x}\right)$ with $x\neq 0$ , as $x$ approaches 0, $\frac{1}{x}$ grows without bound, causing $\sin\left(\frac{1}{x}\right)$ to oscillate rapidly between -1 and 1. The limit $\lim_{x\to 0}p(x)$ doesn't exist because the function doesn't approach any single value. Even though $p(0)=0$ is defined, the non-existent limit creates an oscillating discontinuity. This differs from a jump discontinuity where one-sided limits exist but differ. When a function oscillates infinitely many times near a point without settling on a value, it's an oscillating discontinuity.
For $f(x)=\frac{x^2-9}{x-3}$ for $x\ne3$ and $f(3)=5$ , what type of discontinuity occurs at $x=3$ ?
Answer: Removable discontinuity This problem tests your ability to classify discontinuities by analyzing function behavior at a specific point. The function f(x) = (x² - 9)/(x - 3) can be simplified by factoring the numerator as (x - 3)(x + 3)/(x - 3) = x + 3 for x ≠ 3. This means the limit as x approaches 3 exists and equals 6, but f(3) is defined as 5, which differs from this limit value. Since the limit exists but doesn't equal the function value, this is a removable discontinuity. A jump discontinuity would require different left and right limits, which doesn't occur here since both approach 6. To classify discontinuities, always check: (1) if the limit exists, (2) if the function is defined at that point, and (3) if they're equal.
For $f(x)=\dfrac{x^2-9}{x-3}$ when $x\ne3$ and $f(3)=5$ , what type of discontinuity occurs at $x=3$ ?
Answer: Removable discontinuity Classifying the type of discontinuity in a function is a key skill in understanding where and why a function fails to be continuous. For the function f(x) = (x² - 9)/(x - 3) when x ≠ 3 and f(3) = 5, simplifying the expression gives f(x) = x + 3 for x ≠ 3, so the limit as x approaches 3 is 6. However, since f(3) = 5, which does not equal the limit, the function is discontinuous at x = 3, but the limit exists. This makes it a removable discontinuity because redefining f(3) to 6 would make the function continuous at that point. A tempting distractor might be 'no discontinuity' since the function is defined at x = 3, but it fails because the limit does not match the function value. To classify discontinuities generally, always check if the limit exists and compare it to the function value, then determine if it can be removed by redefinition.
For $f(x)=\frac{x^2-9}{x-3}$ when $x\ne3$ and $f(3)=10$ , what type of discontinuity occurs at $x=3$ ?
Answer: Removable discontinuity Classifying types of discontinuities is essential for analyzing function continuity in AP Calculus BC. For the function f(x), simplifying the expression yields x + 3 for x ≠ 3, so the two-sided limit as x approaches 3 is 6. However, f(3) is defined as 10, which does not equal the limit. This mismatch indicates a removable discontinuity because redefining f(3) to 6 would eliminate the discontinuity and make the function continuous at x = 3. A tempting distractor is no discontinuity, but that fails since the function value does not match the existing limit. A transferable classification strategy is to evaluate limits, compare them to the function value, and determine if redefining the function can achieve continuity.
Let $s(x)=\tan\left(\frac{\pi x}{2}\right)$ . What type of discontinuity occurs at $x=1$ ?
Answer: Infinite discontinuity This problem involves classifying the discontinuity of s(x) = tan(πx/2) at x = 1. When x = 1, we have tan(π/2), which is undefined because cos(π/2) = 0, making tan = sin/cos undefined. As x approaches 1 from the left, πx/2 approaches π/2 from below, and tan(πx/2) approaches +∞. As x approaches 1 from the right, πx/2 approaches π/2 from above, and tan(πx/2) approaches -∞. This behavior—where the function approaches positive infinity from one side and negative infinity from the other—characterizes an infinite discontinuity. A jump discontinuity would require finite one-sided limits. When analyzing trigonometric functions, identify vertical asymptotes where the function is undefined due to division by zero.
For $q(x)=\dfrac{|x|}{x}$ when $x\ne0$ and $q(0)=0$ , what type of discontinuity occurs at $x=0$ ?
Answer: Jump discontinuity This problem requires identifying the discontinuity type for q(x) = |x|/x at x = 0. For x > 0, |x| = x, so q(x) = x/x = 1, giving lim[x→0⁺] q(x) = 1. For x < 0, |x| = -x, so q(x) = -x/x = -1, giving lim[x→0⁻] q(x) = -1. Since the left and right limits exist but are different (and finite), this is a jump discontinuity. The fact that q(0) = 0 is defined doesn't change the discontinuity type. To identify jump discontinuities, check if both one-sided limits exist as finite values but are unequal.
For $q(x)=\frac{|x|}{x}$ when $x\ne 0$ and $q(0)=0$ , what type of discontinuity occurs at $x=0$ ?
Answer: Jump discontinuity This question tests understanding of jump discontinuities in functions with absolute values. The function $q(x)=\frac{|x|}{x}$ equals 1 when $x>0$ and -1 when $x<0$ . Therefore, $\lim_{x\to 0^+}q(x)=1$ and $\lim_{x\to 0^-}q(x)=-1$ . Since these one-sided limits exist but are different, this is a jump discontinuity at $x=0$ . The fact that $q(0)=0$ is defined doesn't change the discontinuity type. A removable discontinuity would require both one-sided limits to be equal. To identify jump discontinuities, check if one-sided limits exist but differ from each other.
Let $q(x)=\frac{|x|}{x}$ for $x\ne0$ and $q(0)=0$ . What type of discontinuity does $q$ have at $x=0$ ?
Answer: Jump discontinuity Classifying types of discontinuities is essential for analyzing function continuity in AP Calculus BC. For q(x) = |x|/x for x ≠ 0 and q(0) = 0, the left-hand limit as x approaches 0 is -1, while the right-hand limit is 1. Both one-sided limits exist but are not equal, creating a sudden jump in the function's values. Additionally, q(0) = 0 matches neither limit, confirming the discontinuity, which is classified as a jump type. A tempting distractor is removable discontinuity, but that fails because the one-sided limits differ, so no single redefinition can make it continuous. A transferable classification strategy is to evaluate one-sided limits; if they exist but differ, it's a jump, whereas agreement might suggest removable or none.
For $u(x)=\dfrac{2x}{x^2-16}$ , what type of discontinuity occurs at $x=4$ ?
Answer: Infinite discontinuity This problem asks about the discontinuity of u(x) = 2x/(x² - 16) at x = 4. First, factor the denominator: x² - 16 = (x - 4)(x + 4), so u(x) = 2x/[(x - 4)(x + 4)]. As x approaches 4, the numerator approaches 8 while the denominator approaches 0, with (x - 4) → 0 and (x + 4) → 8. This causes the function to approach ±∞, creating an infinite discontinuity. The sign depends on the direction of approach: from the left it's -∞, from the right it's +∞. Unlike removable discontinuities, there's no common factor to cancel. Infinite discontinuities occur at vertical asymptotes where denominators approach zero without cancellation.
Consider $p(x)=\sin\!\left(\frac{1}{x}\right)$ for $x\ne0$ and $p(0)=0$ . What discontinuity type occurs at $x=0$ ?
Answer: Oscillating discontinuity Classifying types of discontinuities is essential for analyzing function continuity in AP Calculus BC. For p(x) = sin(1/x) with p(0) = 0, as x approaches 0, 1/x becomes arbitrarily large, causing sin(1/x) to oscillate rapidly between -1 and 1. The limit does not exist because the function values do not approach a single number, instead oscillating indefinitely. This type of behavior is classified as an oscillating discontinuity. A tempting distractor is infinite discontinuity, but that fails because the function remains bounded and does not approach infinity. A transferable classification strategy is to observe if the function oscillates without converging to a limit, distinguishing it from jump or infinite types.
Let $r(x)=\begin{cases}1,&x\ne1\\0,&x=1\end{cases}$ . What type of discontinuity does $r$ have at $x=1$ ?
Answer: Removable discontinuity This problem requires identifying discontinuities in simple piecewise functions. For r(x), the limit as x approaches 1 is lim[x→1] r(x) = 1 (since r(x) = 1 for all x ≠ 1), but r(1) = 0. Because the limit exists and is finite but differs from the function value at that point, this is a removable discontinuity. The discontinuity is not a jump (option B) because both one-sided limits equal 1. To classify as removable, verify that the limit exists finitely but doesn't match the function value.
For $p(x)=\begin{cases}\sin x/x,&x\ne0\\2,&x=0\end{cases}$ , what type of discontinuity occurs at $x=0$ ?
Answer: Removable discontinuity Classifying the type of discontinuity in a function is a key skill in understanding where and why a function fails to be continuous. For p(x) = sin(x)/x when x ≠ 0 and p(0) = 2, the limit as x approaches 0 is 1, which is well-known from calculus. However, p(0) = 2 does not equal this limit, so there is a discontinuity at x = 0, but the limit exists finitely. This is a removable discontinuity since redefining p(0) to 1 would make it continuous. A tempting distractor might be 'no discontinuity' because the function is defined everywhere, but it fails since the limit and function value mismatch. To classify discontinuities generally, always check if the limit exists and compare it to the function value, then determine if it can be removed by redefinition.
Let $r(x)=\begin{cases}x^2,&x\le1\\x^2+2,&x>1\end{cases}$ . What type of discontinuity occurs at $x=1$ ?
Answer: Jump discontinuity Classifying the type of discontinuity in a function is a key skill in understanding where and why a function fails to be continuous. For the piecewise r(x), at x = 1, the left-hand limit is 1 from x², the right-hand limit is 3 from x² + 2, and r(1) = 1. The one-sided limits are finite but unequal, so the limit does not exist, indicating a jump discontinuity. The graph jumps from 1 to 3 across x = 1. A tempting distractor might be 'no discontinuity' since it's defined at x = 1, but it fails because the right limit differs. To classify discontinuities generally, always check if the limit exists and compare it to the function value, then determine if it can be removed by redefinition.
Consider $p(x)=\sin\left(\dfrac{1}{x}\right)$ for $x\ne0$ and $p(0)=0$ . What type of discontinuity is at $x=0$ ?
Answer: Oscillating discontinuity This problem involves classifying the discontinuity of p(x) = sin(1/x) at x = 0. As x approaches 0, the argument 1/x oscillates between -∞ and +∞ increasingly rapidly, causing sin(1/x) to oscillate between -1 and 1 infinitely many times. The limit lim[x→0] sin(1/x) does not exist because the function doesn't approach any single value. Even though p(0) = 0 is defined, the oscillating behavior prevents continuity. This isn't a jump discontinuity because there aren't just two distinct limit values. When a function oscillates infinitely near a point without settling on a limit, it's an oscillating discontinuity.
Let $u(x)=\begin{cases}\ln(x),&x>0\\0,&x\le0\end{cases}$ . What type of discontinuity occurs at $x=0$ ?
Answer: Infinite discontinuity This problem involves analyzing the piecewise function u(x) where u(x) = ln(x) for x > 0 and u(x) = 0 for x ≤ 0. As x approaches 0 from the right, ln(x) approaches -∞ because the natural logarithm of positive numbers approaching zero becomes increasingly negative without bound. From the left, u(x) = 0 for all x ≤ 0, so lim(x→0⁻) u(x) = 0. Since the right-hand limit approaches -∞ while the left-hand limit is finite, this creates an infinite discontinuity at x = 0. A jump discontinuity would require both one-sided limits to be finite. When analyzing logarithmic functions near their domain boundaries, remember that ln(x) → -∞ as x → 0⁺.
Let $t(x)=\begin{cases}x^2,&x\le1\\x+2,&x>1\end{cases}$ . What type of discontinuity does $t$ have at $x=1$ ?
Answer: Jump discontinuity This problem requires classifying the discontinuity of the piecewise function t(x) at x = 1. From the left: lim[x→1⁻] t(x) = lim[x→1⁻] x² = 1² = 1, and t(1) = 1² = 1. From the right: lim[x→1⁺] t(x) = lim[x→1⁺] (x + 2) = 1 + 2 = 3. Since the left-hand limit (1) differs from the right-hand limit (3), and both are finite, this is a jump discontinuity. The function "jumps" from y = 1 to y = 3 as x passes through 1. For piecewise functions, always calculate both one-sided limits to detect jumps.
A piecewise function is $h(x)=\begin{cases}2x+1,&x<0\\3-x,&x\ge0\end{cases}$ . What discontinuity type occurs at $x=0$ ?
Answer: Jump discontinuity Classifying the type of discontinuity in a function is a key skill in understanding where and why a function fails to be continuous. For the piecewise function h(x), the left-hand limit at x = 0 is 1 from 2x + 1, while the right-hand limit is 3 from 3 - x, and h(0) = 3. Since the one-sided limits exist but are not equal, the overall limit does not exist, indicating a jump discontinuity. The function 'jumps' from approaching 1 on the left to 3 on the right. A tempting distractor might be 'no discontinuity' because the function is defined at x = 0, but it fails as the limits from left and right differ. To classify discontinuities generally, always check if the limit exists and compare it to the function value, then determine if it can be removed by redefinition.
For $h(x)=\frac{1}{(x+2)^2}$ , identify the type of discontinuity of $h$ at $x=-2$ .
Answer: Infinite discontinuity Classifying types of discontinuities is essential for analyzing function continuity in AP Calculus BC. For h(x) = 1/ $(x + 2)^2$ , the function is undefined at x = -2 because the denominator is zero. As x approaches -2 from either side, the denominator approaches 0 from the positive side, causing h(x) to approach +∞. This unbounded behavior classifies the discontinuity as infinite, characterized by a vertical asymptote. A tempting distractor is removable discontinuity, but that fails because the limit does not exist finitely, as the function values grow without bound. A transferable classification strategy is to check if the function approaches ±∞ near the point, indicating an infinite discontinuity, after confirming the limit does not exist finitely.
For $h(x)=\dfrac{1}{(x-2)^2}$ , what type of discontinuity occurs at $x=2$ ?
Answer: Infinite discontinuity This problem asks us to identify the discontinuity type for h(x) = 1/(x - 2)² at x = 2. As x approaches 2 from either side, the denominator (x - 2)² approaches 0 while remaining positive, causing the function to approach +∞. Since lim[x→2] h(x) = +∞, this is an infinite discontinuity. The function cannot be made continuous by redefining h(2) because the limit doesn't exist as a finite value. It's not removable because we can't "fill the hole" with any finite value. To identify infinite discontinuities, look for vertical asymptotes where the function approaches ±∞.
Let $q(x)=\frac{|x|}{x}$ for $x\ne0$ and $q(0)=0$ . What type of discontinuity occurs at $x=0$ ?
Answer: Jump discontinuity This problem involves analyzing q(x) = |x|/x for x ≠ 0, with q(0) = 0. For x > 0, we have |x| = x, so q(x) = x/x = 1, giving lim(x→0⁺) q(x) = 1. For x < 0, we have |x| = -x, so q(x) = -x/x = -1, giving lim(x→0⁻) q(x) = -1. Since the left and right limits exist but differ (1 ≠ -1), this creates a jump discontinuity at x = 0. The fact that q(0) is defined as 0 doesn't change the classification—jump discontinuities depend only on the one-sided limits being different. An oscillating discontinuity would require the function to oscillate without approaching any limit. When working with absolute value functions, always consider the behavior from both positive and negative directions separately.

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