Defining Continuity at a Point - AP Calculus BC

1. For $p(x)=\begin{cases}\sin x/x,&x\ne0\\1,&x=0\end{cases}$, is $p$ continuous at $x=0$, and why?
   Answer: Yes; $p(0)$ exists, $\lim_{x\to0}p(x)$ exists, and $\lim_{x\to0}p(x)=p(0)$. This question tests continuity of a function with a removable discontinuity that has been properly "filled in." For continuity at x = 0, we need p(0) to exist, lim[x→0] p(x) to exist, and these to be equal. From the definition, p(0) = 1. For x ≠ 0, p(x) = sin(x)/x, and the famous limit lim[x→0] sin(x)/x = 1. Since p(0) = 1 and lim[x→0] p(x) = 1, all three conditions for continuity are satisfied: the function value exists, the limit exists, and they're equal. Choice E might tempt students who compute the limit incorrectly, but this is a classic example of properly defining a function at a removable discontinuity. Remember that sin(x)/x → 1 as x → 0 is a fundamental limit worth memorizing.
2. Suppose $r(x)=\frac{1}{x-5}$. Is $r$ continuous at $x=5$, and why?
   Answer: No; $\lim_{x\to5}r(x)$ does not exist as a finite number and $r(5)$ is undefined. This question checks continuity at a point for a rational function with a vertical asymptote. For r to be continuous at x=5, r(5) must be defined, the limit as x approaches 5 must exist finitely, and they must equal. Here, r(5) is undefined due to division by zero, and the limit is infinite from both sides, so neither the value nor a finite limit exists. This is an example of an infinite discontinuity. A distractor might claim the limit exists but does not equal r(5), but the limit is not finite and r(5) is undefined. 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).
3. Let $q(x)=\frac{x^2}{x}$ for $x\ne0$ and $q(0)=0$. Is $q$ continuous at $x=0$, and why?
   Answer: Yes; $q(0)$ is defined, $\lim_{x\to0}q(x)$ exists, and equals $q(0)$. 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.
4. Let $h(x)=\begin{cases}x+1,&x<0\\x^2+1,&x\ge0\end{cases}$. Is $h$ continuous at $x=0$, and why?
   Answer: Yes; $\lim_{x\to0}h(x)$ exists and equals $h(0)$. This question tests continuity of a piecewise function at the transition point. For h to be continuous at x = 0, we need h(0) defined, lim[x→0] h(x) to exist, and these to be equal. From the definition, h(0) = 0² + 1 = 1 (using the x ≥ 0 formula). For the limit, we check both sides: lim[x→0⁻] h(x) = lim[x→0⁻] (x + 1) = 0 + 1 = 1, and lim[x→0⁺] h(x) = lim[x→0⁺] (x² + 1) = 0² + 1 = 1. Since both one-sided limits equal 1, lim[x→0] h(x) = 1 = h(0), so h is continuous at x = 0. Choice D incorrectly claims the limit doesn't equal h(0), when both equal 1. For piecewise functions, always check both one-sided limits at transition points and verify they match the function value there.
5. For $t(x)=\tan x$, is $t$ continuous at $x=\dfrac{\pi}{2}$, and why?
   Answer: No; $t(\pi/2)$ is undefined, so continuity fails. This question examines continuity of the tangent function at a point where it has a vertical asymptote. For continuity at x = π/2, we need t(π/2) to exist, lim[x→π/2] t(x) to exist, and these to be equal. The tangent function tan(x) = sin(x)/cos(x), and since cos(π/2) = 0, we have division by zero, making t(π/2) undefined. This immediately violates the first condition for continuity. Choice E might seem correct because tan(x) → +∞ as x → π/2⁻ and tan(x) → -∞ as x → π/2⁺, suggesting the limit doesn't exist, but the primary issue is that the function itself is undefined at π/2. Remember that trigonometric functions have discontinuities at their vertical asymptotes where they're undefined.
6. For $g(x)=\dfrac{x^2-4}{x-2}$, is $g$ continuous at $x=2$, and why?
   Answer: No; $g(2)$ is undefined, so continuity fails. 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.
7. For $s(x)=\begin{cases}\frac{\sin x}{x},&x\ne0\\1,&x=0\end{cases}$, is $s$ continuous at $x=0$, and why?
   Answer: Yes; $s(0)$ exists and equals $\lim_{x\to0}s(x)$. This question explores continuity at a point using the sinc function. For s to be continuous at x=0, s(0) must be defined, the limit as x approaches 0 must exist, and they must be equal. The limit of (sin x)/x as x approaches 0 is 1 by standard limit rules, and s(0)=1, so conditions hold, making s continuous. This removable discontinuity is fixed by defining s(0)=1. A common error is thinking the limit does not exist due to oscillation of sin x, but the division by x squeezes it to 1. 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).
8. For $r(x)=\begin{cases}x^2,&x\ne-1\\0,&x=-1\end{cases}$, is $r$ continuous at $x=-1$, and why?
   Answer: No; $\lim_{x\to-1}r(x)\ne r(-1)$, so continuity fails. This question presents a function with a point discontinuity where the function value differs from what the formula suggests. For continuity at x = -1, we need r(-1) to exist, lim[x→-1] r(x) to exist, and these to be equal. From the definition, r(-1) = 0. For x ≠ -1, r(x) = x², so lim[x→-1] r(x) = lim[x→-1] x² = (-1)² = 1. Since r(-1) = 0 and lim[x→-1] r(x) = 1, we have 0 ≠ 1, violating the third continuity condition. Choice D might attract students who think the limit doesn't exist because of the piecewise definition, but the limit of x² as x approaches -1 clearly exists. When a function is redefined at a single point, the limit still follows the original formula.
9. Let $v(x)=\begin{cases}x^3,&x\ne -2\\-8,&x=-2\end{cases}$. Is $v$ continuous at $x=-2$, and why?
   Answer: Yes; $v(-2)$ exists, $\lim_{x\to-2}v(x)$ exists, and they are equal. This question asks about continuity of a polynomial that's been redefined at a single point. For continuity at x = -2, we need v(-2) to exist, lim[x→-2] v(x) to exist, and these to be equal. We're given v(-2) = -8, and for x ≠ -2, v(x) = x³, so lim[x→-2] v(x) = lim[x→-2] x³ = (-2)³ = -8. Since v(-2) = -8 = lim[x→-2] v(x), all three continuity conditions are satisfied. This demonstrates that redefining a continuous function at a single point to match its original limit value maintains continuity. Students might think any piecewise definition creates discontinuity, but matching the limit preserves continuity. For polynomial-based piecewise functions: calculate the polynomial's limit, then check if the special case matches.
10. For $f(x)=\frac{x^2-9}{x-3}$ when $x\ne3$ and $f(3)=5$, is $f$ continuous at $x=3$, and why?
    Answer: No; $\lim_{x\to3}f(x)$ exists but is not equal to $f(3)$. This question tests the definition of continuity at a point. For a function f to be continuous at x=3, three conditions must hold: f(3) must be defined, the limit as x approaches 3 must exist, and the limit must equal f(3). Here, f(3) is defined as 5, and by simplifying the expression to x+3 for x≠3, the limit as x approaches 3 is 6, which exists. However, since 6 does not equal 5, the function is not continuous at x=3. A tempting distractor is choice C, which claims the limit does not exist due to the zero denominator, but this ignores the removable discontinuity after simplification. To check continuity at a point, always use this checklist: verify the function value is defined, confirm the limit exists, and ensure they match.
11. Let $g(x)=\begin{cases}x+2,&x<1\\4,&x=1\\x^2+3,&x>1\end{cases}$. Is $g$ continuous at $x=1$, and why?
    Answer: No; the one-sided limits at $x=1$ are not equal, so the limit does not exist. 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.
12. Let $t(x)=\begin{cases}x^2,&x\le2\\4,&x>2\end{cases}$. Is $t$ continuous at $x=2$, and why?
    Answer: Yes; $\lim_{x\to2}t(x)$ exists and equals $t(2)$. This question tests continuity of a piecewise function where the pieces meet at the transition point. For continuity at x = 2, we need t(2) defined, lim[x→2] t(x) to exist, and these to be equal. From the x ≤ 2 piece, t(2) = 2² = 4. For the limit: lim[x→2⁻] t(x) = lim[x→2⁻] x² = 4, and lim[x→2⁺] t(x) = lim[x→2⁺] 4 = 4. Since both one-sided limits equal 4, lim[x→2] t(x) = 4 = t(2), satisfying all continuity conditions. Choice B incorrectly claims the limit doesn't equal t(2), when both equal 4. For piecewise functions, pay attention to which piece defines the function value at the transition point (here, x ≤ 2 includes x = 2).
13. For $v(x)=\dfrac{1}{x-4}+2$, is $v$ continuous at $x=4$, and why?
    Answer: No; $v(4)$ is undefined, so continuity fails. This question tests continuity of a rational function at a point where it has a vertical asymptote. For continuity at x = 4, we need v(4) to exist, lim[x→4] v(x) to exist, and these to be equal. The function v(x) = 1/(x-4) + 2 has a denominator that equals zero when x = 4, making v(4) undefined due to division by zero. This immediately fails the first condition for continuity. Choice D might attract students who focus on the limit behavior (as x → 4⁺, v(x) → +∞ and as x → 4⁻, v(x) → -∞), but the fundamental issue is that v(4) doesn't exist. For rational functions, always check where denominators equal zero—these points are automatically discontinuous.
14. Let $u(x)=\begin{cases}\sin x,&x\ne\pi\\0,&x=\pi\end{cases}$. Is $u$ continuous at $x=\pi$, and why?
    Answer: Yes; $\lim_{x\to\pi}u(x)$ exists and equals $u(\pi)$. This question tests the definition of continuity at a point. For u to be continuous at x=π, u(π) must be defined, the limit as x approaches π must exist, and they must be equal. Here, u(π)=0 is defined, and the limit of sin(x) as x approaches π is sin(π)=0, which matches. Thus, u is continuous at x=π despite the piecewise definition. A tempting distractor is choice C, which claims the limit does not equal u(π), but both are 0. To check continuity at a point, always use this checklist: verify the function value is defined, confirm the limit exists, and ensure they match.
15. Let $u(x)=\begin{cases}\frac{x^2-1}{x-1},&x\ne1\\3,&x=1\end{cases}$. Is $u$ continuous at $x=1$, and why?
    Answer: No; $\lim_{x\to1}u(x)$ exists but does not equal $u(1)$. This question involves a removable discontinuity where the function has been redefined at the problem point. For $u(x) = \frac{x^2-1}{x-1}$ when $x \neq 1$, we can factor and simplify: $u(x) = \frac{(x-1)(x+1)}{x-1} = x+1$ for $x \neq 1$. Therefore, $\lim_{x \to 1} u(x) = \lim_{x \to 1} (x+1) = 2$. However, $u(1) = 3$ by definition, so while the limit exists, it doesn't equal the function value: $\lim_{x \to 1} u(x) = 2 \neq 3 = u(1)$. Students might think that defining the function at the discontinuity automatically makes it continuous, but the value must match the limit. To ensure continuity at removable discontinuities: calculate the limit of the simplified form and define the function value to match.
16. Let $p(x)=\begin{cases}\sin(1/x),&x\ne0\\0,&x=0\end{cases}$. Is $p$ continuous at $x=0$, and why?
    Answer: No; $p(0)$ exists but $\lim_{x\to0}p(x)$ does not exist. This question involves a function with oscillatory behavior near the point of interest. For continuity at x = 0, we need p(0) defined, lim[x→0] p(x) to exist, and these to be equal. We have p(0) = 0 (given), but for x ≠ 0, p(x) = sin(1/x) oscillates rapidly between -1 and 1 as x approaches 0. As x → 0⁺, 1/x → +∞ and sin(1/x) oscillates without approaching any single value; similarly as x → 0⁻, 1/x → -∞ and sin(1/x) oscillates. Therefore, lim[x→0] p(x) does not exist. Choice D incorrectly suggests the limit exists but doesn't equal p(0)—the limit doesn't exist at all due to oscillation. When dealing with compositions involving 1/x, check for unbounded behavior or oscillation that prevents the limit from existing.
17. Let $s(x)=\begin{cases}2x,&x\le4\\x+4,&x>4\end{cases}$. Is $s$ continuous at $x=4$, and why?
    Answer: Yes; $s(4)$ is defined, $\lim_{x\to4}s(x)$ exists, and equals $s(4)$. This question examines continuity at the transition point of a piecewise function. For $s(x)$, we need to check continuity at $x = 4$. From the definition, $s(4) = 2(4) = 8$ (using the first piece since $x \leq 4$). For the limit: $\lim_{x \to 4^-} s(x) = \lim_{x \to 4^-} 2x = 8$ and $\lim_{x \to 4^+} s(x) = \lim_{x \to 4^+} (x+4) = 8$. Since both one-sided limits equal 8, we have $\lim_{x \to 4} s(x) = 8 = s(4)$. Students might incorrectly evaluate $s(4)$ using the wrong piece of the function. For piecewise functions: carefully note which piece applies at the transition point (check ≤ vs < ), then verify both one-sided limits match this value.
18. For $p(x)=\begin{cases}\sin x,&x\ne0\\1,&x=0\end{cases}$, is $p$ continuous at $x=0$, and why?
    Answer: No; $p(0)$ exists but $\lim_{x\to0}p(x)\ne p(0)$. This question examines continuity at a point for a piecewise function involving sine. Continuity at x=0 requires p(0) to be defined, the limit as x approaches 0 to exist, and equality between them. Here, p(0)=1 is defined, and the limit of sin x as x approaches 0 is 0, but 0 ≠ 1, so continuity fails. This illustrates a removable discontinuity if redefined, but as given, it is discontinuous. A distractor might claim the limit does not exist due to oscillation, but sin x approaches 0 without issue. 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).
19. For $s(x)=\sqrt{x-1}$, is $s$ continuous at $x=1$, and why?
    Answer: Yes; $s(1)$ exists, $\lim_{x\to1}s(x)$ exists, and $\lim_{x\to1}s(x)=s(1)$. This question tests continuity of a square root function at the edge of its domain. For continuity at x = 1, we need s(1) to exist, lim[x→1] s(x) to exist, and these to be equal. First, s(1) = √(1-1) = √0 = 0, so the function value exists. For the limit, as x approaches 1 from the right, √(x-1) approaches √0 = 0. Note that we can only approach from the right since √(x-1) requires x ≥ 1. The one-sided limit lim[x→1⁺] s(x) = 0 equals s(1) = 0. Choice B might tempt students who think s(1) is undefined because of the square root, but √0 = 0 is well-defined. For functions with restricted domains, continuity at boundary points only requires the appropriate one-sided limit to exist and equal the function value.
20. Suppose $v(x)=\frac{x^2-1}{x-1}$ for $x\ne1$ and $v(1)$ is not defined. Is $v$ continuous at $x=1$, and why?
    Answer: No; $v(1)$ is undefined, so continuity fails. This question tests the definition of continuity at a point. For v to be continuous at x=1, v(1) must be defined, the limit as x approaches 1 must exist, and they must be equal. Here, v(1) is explicitly not defined, failing the first condition, even though the limit exists as 2 after simplification. Therefore, v is not continuous at x=1. A tempting distractor is choice A, which claims the limit's existence implies continuity, but the undefined function value prevents it. To check continuity at a point, always use this checklist: verify the function value is defined, confirm the limit exists, and ensure they match.

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