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.

This question tests understanding of continuity when a function value is redefined at a point. The function $p(x) = \sin x$ for $x \neq 0$ has $\lim_{x \to 0} p(x) = \lim_{x \to 0} \sin x = \sin(0) = 0$. However, we're told that $p(0) = 1$. Since $p(0)$ is defined as 1, the limit exists and equals 0, but $\lim_{x \to 0} p(x) = 0 \neq 1 = p(0)$, the function is not continuous at $x = 0$. Students might assume that since sine is normally continuous everywhere, this function must be continuous, but redefining the value at a single point can break continuity. To verify continuity: always check that the limit equals the actual function value, not what it "should" be.

This question examines continuity at a point where the function has different behaviors from each side. The function $h(x) = \frac{|x|}{x}$ equals 1 when $x > 0$ and -1 when $x < 0$. We have $h(0) = 0$ by definition. For the limit: $\lim_{x \to 0^+} h(x) = 1$ and $\lim_{x \to 0^-} h(x) = -1$. Since these one-sided limits are different, $\lim_{x \to 0} h(x)$ does not exist. A student might think that defining $h(0) = 0$ makes the function continuous, but the limit must exist independently of how we define the function at the point. When checking continuity involving absolute values or sign functions: always evaluate both one-sided limits separately.

This question tests the definition of continuity at a point. For p to be continuous at x=0, p(0) must be defined, the limit as x approaches 0 must exist, and they must be equal. Here, p(0)=1 is defined, but the left-hand limit is -1 and the right-hand limit is 1, so the two-sided limit does not exist. Therefore, p is not continuous at x=0. A tempting distractor is choice A, which claims the limit equals p(0), but the differing one-sided limits prevent the limit from existing. To check continuity at a point, always use this checklist: verify the function value is defined, confirm the limit exists, and ensure they match.

This question tests the definition of continuity at a point. For s to be continuous at x=-1, s(-1) must be defined, the limit as x approaches -1 must exist, and they must be equal. Here, s(-1)=3 is defined, and the limit is 1 (from x²), which exists but does not equal 3. Therefore, s is not continuous at x=-1. A tempting distractor is choice A, which claims continuity because the limit equals s(-1), but 1≠3. To check continuity at a point, always use this checklist: verify the function value is defined, confirm the limit exists, and ensure they match.

This question tests the definition of continuity at a point. For g to be continuous at x=1, g(1) must be defined, the limit as x approaches 1 must exist, and they must be equal. Here, g(1)=4 is defined, but the left-hand limit is 3 and the right-hand limit is 4, so the two-sided limit does not exist. Therefore, g is not continuous at x=1. A tempting distractor is choice B, which assumes both one-sided limits equal g(1), but the left limit differs. To check continuity at a point, always use this checklist: verify the function value is defined, confirm the limit exists, and ensure they match.

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 an endpoint of the domain for a square root function. For h to be continuous at x=3, where the domain starts, h(3) must be defined and equal the right-hand limit as x approaches 3 from above, since the left-hand limit is not applicable. Here, h(3)=0, and the right-hand limit is also 0, satisfying the conditions for continuity at the endpoint. In calculus, functions like square root are considered continuous on their closed domain intervals at the endpoints. A tempting distractor might argue the two-sided limit does not exist due to the domain, but continuity at endpoints uses the appropriate one-sided limit. 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 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).