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 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.

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.

Classifying the type of discontinuity in a function is a key skill in understanding where and why a function fails to be continuous. For s(x) = |x|/x when x ≠ 0 and s(0) = 0, the left-hand limit is -1 and the right-hand limit is 1, while s(0) = 0. Since the one-sided limits exist but are not equal, the overall limit does not exist, characterizing a jump discontinuity. The function jumps from -1 to 1 across x = 0, with the defined value in between not affecting the classification. A tempting distractor might be 'removable discontinuity' because it's defined at 0, but it fails as no single redefinition can match both sides. 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.

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.

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.

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.

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 ±∞.