This problem involves a rational function with a removable discontinuity that has been "filled in" at x = 3. For continuity at x = 3, we need s(3) to exist, lim[x→3] s(x) to exist, and these to be equal. We're given s(3) = 6, and for x ≠ 3, we can simplify: s(x) = (x² - 9)/(x - 3) = (x + 3)(x - 3)/(x - 3) = x + 3. Therefore, lim[x→3] s(x) = lim[x→3] (x + 3) = 3 + 3 = 6. Since s(3) = 6 = lim[x→3] s(x), all three continuity conditions are satisfied. The key insight is recognizing that defining the function value to match the limit of the simplified expression creates continuity. For removable discontinuities: factor and simplify first, find the limit, then verify the defined value matches.

This question tests whether a function is continuous at a point by checking if all three conditions are satisfied. For continuity at x = 1, we need: (1) f(1) to exist, (2) lim[x→1] f(x) to exist, and (3) these two values to be equal. Here, f(1) = 3 (given by the piecewise definition), and lim[x→1] f(x) = lim[x→1] (x² - 1) = 1² - 1 = 0. Since f(1) = 3 ≠ 0 = lim[x→1] f(x), the function fails the third condition for continuity. A common mistake is assuming that having a defined value at a point guarantees continuity, but all three conditions must be met. To check continuity: verify the function value exists, calculate the limit, then compare—if any step fails or they're unequal, the function is discontinuous.

This question evaluates continuity at a point for a modified quadratic function. Continuity at x=-2 requires v(-2) to be defined, the limit as x approaches -2 to exist, and equality. v(-2)=0 is defined, the limit is $(-2)^2$=4 from both sides, but 4 ≠ 0, so it fails. This is a removable discontinuity. A wrong choice might say the limit does not exist, but it does exist as 4. To check continuity at a point, use this checklist: verify f(a) is defined, confirm the limit exists by checking one-sided limits match, and ensure they equal f(a).

This question tests whether redefining a continuous function at one point maintains continuity. The function $v(x) = \cos x$ for $x \neq \pi$ normally has $\cos(\pi) = -1$. Since $\lim_{x \to \pi} v(x) = \lim_{x \to \pi} \cos x = \cos(\pi) = -1$, and we're told $v(\pi) = -1$, all three continuity conditions are satisfied: $v(\pi)$ is defined, the limit exists and equals -1, and this equals $v(\pi)$. A student might think that treating $x = \pi$ as a special case breaks continuity, but since the redefined value matches what the function would naturally be, continuity is preserved. When a piecewise definition matches the original function's value, continuity is maintained.

This question involves a square root function at the edge of its domain, testing continuity where the function begins. For continuity at x = 1, we need u(1) defined, lim[x→1] u(x) to exist, and these to be equal. Since u(x) = √(x - 1) is defined for x ≥ 1, we have u(1) = √(1 - 1) = √0 = 0. For the limit, we only consider the right-hand limit since the function isn't defined for x < 1: lim[x→1⁺] u(x) = lim[x→1⁺] √(x - 1) = √0 = 0. Since this one-sided limit equals u(1), the function is continuous at x = 1. Choice A incorrectly suggests the limit doesn't equal u(1), but both equal 0. For functions with restricted domains, continuity at boundary points only requires checking the limit from within the domain.

This question tests recognition of when a function cannot be made continuous at a point. The function $r(x) = \frac{1}{x-2}$ for $x \neq 2$ has a vertical asymptote at $x = 2$. As $x$ approaches 2 from the right, $r(x) \to +\infty$, and as $x$ approaches 2 from the left, $r(x) \to -\infty$. Most importantly, $r(2)$ is not defined in the problem statement. Students might think they could define $r(2)$ to make it continuous, but no finite value would work since the limit doesn't exist. For rational functions with non-removable discontinuities: check if the denominator's zero persists after simplification - if yes, continuity is impossible.

This question tests whether a function is continuous at a point by checking the three conditions for continuity. For continuity at x = 1, we need: (1) f(1) to be defined, (2) lim[x→1] f(x) to exist, and (3) these two values to be equal. Here, f(1) = 3 (defined), and for x ≠ 1, f(x) = x² - 1, so lim[x→1] f(x) = lim[x→1] (x² - 1) = 1² - 1 = 0. Since lim[x→1] f(x) = 0 ≠ 3 = f(1), the function fails the third condition for continuity. Choice A incorrectly claims the limit equals f(1), when actually 0 ≠ 3. To check continuity at a point: verify the function is defined there, calculate the limit, then compare these two values—all three conditions must be satisfied.

This question examines continuity at a point where a piecewise function changes its formula. For continuity at x = 2, we need q(2) to exist, lim[x→2] q(x) to exist, and these to be equal. Since 2 satisfies x ≤ 2, we have q(2) = 2 + 3 = 5. For the limit, we check: lim[x→2⁻] q(x) = lim[x→2⁻] (x + 3) = 5 and lim[x→2⁺] q(x) = lim[x→2⁺] 7 = 7. Since 5 ≠ 7, the one-sided limits differ, so lim[x→2] q(x) doesn't exist. Choice E might seem plausible if you incorrectly think the limit exists, but when one-sided limits differ, the two-sided limit doesn't exist. At piecewise transition points, always verify both one-sided limits match before concluding the limit exists.

This question asks about continuity at a point where the function has a removable discontinuity. For continuity at x = 2, we need g(2) to exist, lim[x→2] g(x) to exist, and these values to be equal. The function g(x) = (x² - 4)/(x - 2) is undefined at x = 2 because we get 0/0, so g(2) doesn't exist. Even though we could simplify to g(x) = (x + 2)(x - 2)/(x - 2) = x + 2 for x ≠ 2, giving lim[x→2] g(x) = 4, the function value g(2) remains undefined. Choice C might seem correct since the limit does exist, but the first failure point is that g(2) is undefined. Check conditions in order: if the function value doesn't exist, continuity immediately fails.

This question involves a function with a removable discontinuity that has been properly addressed. For $q(x) = \frac{x^2}{x}$ when $x \neq 0$, we can simplify to $q(x) = x$ for $x \neq 0$. Therefore, $\lim_{x \to 0} q(x) = \lim_{x \to 0} x = 0$. We're given that $q(0) = 0$, so all three continuity conditions are satisfied: $q(0)$ is defined, $\lim_{x \to 0} q(x)$ exists and equals 0, and this equals $q(0)$. A common error would be to think the original fraction form prevents continuity at $x = 0$, but the simplified form shows the limit exists. When dealing with rational functions: simplify first, then evaluate limits using the simplified form.