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.

This question tests whether you can identify when a piecewise function has no discontinuity. For x ≠ 1, t(x) = (x² - 1)/(x - 1) = (x - 1)(x + 1)/(x - 1) = x + 1, so lim(x→1) t(x) = 1 + 1 = 2. The function is defined at x = 1 with t(1) = 2, which equals the limit. Since the limit exists, the function is defined at that point, and these values are equal, the function is continuous at x = 1—there is no discontinuity. A removable discontinuity would occur if t(1) were defined differently from the limit value. To verify continuity, always check three conditions: (1) the limit exists, (2) the function is defined, and (3) the limit equals the function value.

This problem involves classifying the discontinuity of h(x) = 1/(x + 2)² at x = -2. As x approaches -2 from either direction, the denominator (x + 2)² approaches 0 while remaining positive (since it's squared), causing the function to approach positive infinity. This behavior—where the function grows without bound as x approaches the discontinuity from both sides—defines an infinite discontinuity. A jump discontinuity would require finite but different one-sided limits, which doesn't occur here. To identify infinite discontinuities, look for denominators that approach zero without cancellation, particularly when the function approaches ±∞ from at least one side.

This question tests recognition of oscillating discontinuities, which occur when a function oscillates infinitely as it approaches a point. For p(x) = sin(1/x), as x approaches 0, the argument 1/x grows without bound, causing sin(1/x) to oscillate rapidly between -1 and 1. This means the limit as x approaches 0 does not exist—not because the function approaches infinity, but because it oscillates without settling on any value. Even though p(0) is defined as 0, the non-existence of the limit due to oscillation creates an oscillating discontinuity. A removable discontinuity would require the limit to exist, which it doesn't here. To identify oscillating discontinuities, look for compositions involving periodic functions with arguments that approach infinity.

This problem involves analyzing v(x) = cos(x)/x at x = 0. As x approaches 0, cos(x) approaches cos(0) = 1, while the denominator x approaches 0. This gives a form of 1/0, which means the function approaches ±∞ depending on the sign of x. From the right (x → 0⁺), v(x) → +∞, and from the left (x → 0⁻), v(x) → -∞. This is an infinite discontinuity, not removable, because the limit is infinite rather than finite. The fact that v(0) = 1 is defined doesn't change the discontinuity type. When a continuous numerator is divided by x at x = 0, expect an 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.

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 ±∞.

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.

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.

Classifying the type of discontinuity in a function is a key skill in understanding where and why a function fails to be continuous. For u(x) = (x - 4)/(√x - 2) for x ≥ 0 and u(4) = 7, simplifying by rationalizing gives u(x) = √x + 2 for x ≠ 4, so the limit as x approaches 4 is 4. However, u(4) = 7 ≠ 4, creating a discontinuity where the limit exists finitely. This is removable since redefining u(4) to 4 would ensure continuity. A tempting distractor might be 'infinite discontinuity' before simplifying, but it fails because the indeterminate form resolves to a finite limit. 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.