Defining Continuity at a Point - AP Calculus AB

1. Let $N(x)=\begin{cases}\frac{x-4}{|x-4|},&x\ne4\\0,&x=4\end{cases}$. Is $N$ continuous at $x=4$, and why?
   Answer: No; $\lim_{x\to4^-}N(x)=-1$ and $\lim_{x\to4^+}N(x)=1$, so the limit does not exist. Continuity at x=4: N(4)=0, left -1, right 1, DNE, not continuous. Sign jump. Omission: not evaluating sides. Jump. Strategy: (1) f(a)? (2) Limit? (3) Match?
2. Suppose $h(0)=1$, $\lim_{x\to0^-}h(x)=1$, and $\lim_{x\to0^+}h(x)=2$. Is $h$ continuous at $0$, and why?​​
   Answer: No; $\lim_{x\to0}h(x)$ does not exist since the one-sided limits are different, so it cannot equal $h(0)$. For continuity at x = 0, we need: (1) h(0) exists, (2) lim[x→0] h(x) exists, and (3) they are equal. We're given h(0) = 1, so condition (1) is satisfied. However, for the limit to exist, the left and right limits must be equal: lim[x→0⁻] h(x) = 1 and lim[x→0⁺] h(x) = 2. Since 1 ≠ 2, the two-sided limit does not exist, violating condition (2). When the limit doesn't exist, it cannot equal h(0), so h is not continuous at 0. A common error is thinking that having both one-sided limits guarantees the two-sided limit exists—they must be equal. Continuity checklist: ✓ Function defined at point, ✗ Limit exists, ✗ Limit equals function value.
3. Let $p(x)=\begin{cases}\dfrac{\sin x}{x},&x\ne0\\1,&x=0\end{cases}$. Is $p$ continuous at $x=0$, and why?​
   Answer: Yes; $\lim_{x\to0}p(x)=1$ and $p(0)=1$, so the limit exists and equals the value. To check continuity at x = 0, we verify: (1) p(0) exists, (2) lim[x→0] p(x) exists, and (3) they are equal. From the definition, p(0) = 1, satisfying condition (1). For the limit, we need lim[x→0] (sin x)/x, which is a famous limit equal to 1 (can be proven using L'Hôpital's rule or the squeeze theorem). Since lim[x→0] p(x) = 1 and p(0) = 1, all three conditions are met—p is continuous at x = 0. Students often assume sin x/x is undefined at 0, but the limit exists and the piecewise definition assigns the correct value. Continuity checklist: ✓ Function defined at point, ✓ Limit exists, ✓ Limit equals function value.
4. For $c(x)=\begin{cases}1,&x<1\\2,&x\ge1\end{cases}$, is $c$ continuous at $x=1$, and why?
   Answer: No; $\lim_{x\to1^-}c(x)=1$ and $\lim_{x\to1^+}c(x)=2$, so $\lim_{x\to1}c(x)$ does not exist. Continuity requires the function value, existing limit, and their match at the point. For c(x) at x = 1, c(1) = 2, but left limit = 1 and right = 2, so limit does not exist, discontinuous. A typical omission is not checking if one-sided limits agree. This is a step function discontinuity. Constants on intervals need to match for continuity. To check continuity at any point a, use this checklist: ensure f(a) is defined, compute $\lim_{x→a}$ f(x) and confirm it exists, then check if $\lim_{x→a}$ f(x) = f(a).
5. Let $t(x)=\begin{cases}x+2,&x\le -1\\x^2,&x>-1\end{cases}$. Is $t$ continuous at $x=-1$, and why?​
   Answer: Yes; $\lim_{x\to-1}t(x)=1$ and $t(-1)=1$, so the limit exists and equals the value. To check continuity at x = -1, we need: (1) t(-1) exists, (2) lim[x→-1] t(x) exists, and (3) they are equal. Since -1 is included in the first piece (x ≤ -1), we have t(-1) = -1 + 2 = 1, satisfying condition (1). For the limit: lim[x→-1⁻] t(x) = lim[x→-1⁻] (x + 2) = -1 + 2 = 1, and lim[x→-1⁺] t(x) = lim[x→-1⁺] (x²) = (-1)² = 1. Both one-sided limits equal 1, so lim[x→-1] t(x) = 1. Since t(-1) = 1 and the limit = 1, all conditions are met—t is continuous at x = -1. Students often assume different formulas mean discontinuity, but matching values at the transition point ensures continuity. Continuity checklist: ✓ Function defined at point, ✓ Limit exists, ✓ Limit equals function value.
6. Let $u(x)=\begin{cases}\frac{x^2-1}{x-1},&x\ne1\\2,&x=1\end{cases}$. Is $u$ continuous at $x=1$, and why?
   Answer: Yes; $\lim_{x\to1}u(x)=2$ exists and equals $u(1)=2$. For continuity at x = 1, we need u(1) defined, lim[x→1] u(x) to exist, and these values to match. The function defines u(1) = 2, satisfying condition 1. To find the limit, we simplify the rational expression: lim[x→1] (x² - 1)/(x - 1) = lim[x→1] (x + 1)(x - 1)/(x - 1) = lim[x→1] (x + 1) = 2, so the limit exists and equals 2. Since lim[x→1] u(x) = 2 = u(1), all continuity conditions are satisfied. This shows how removable discontinuities can be "removed" by defining the function value to equal the limit. Continuity checklist: (1) u(1) = 2 ✓, (2) Limit = 2 after factoring ✓, (3) They match ✓.
7. Let $u(x)=\begin{cases}x^2+4x+4,&x\ne-2\\0,&x=-2\end{cases}$. Is $u$ continuous at $x=-2$, and why?
   Answer: Yes; $\lim_{x\to-2}u(x)=0$ and $u(-2)=0$. For continuity at x = a, ensure the three conditions. At x = -2 for u(x), u(-2) = 0, limit (x+2)² → 0, matches, continuous. Common mistake: assuming quadratic form implies no limit without factoring. It factors to a hole filled correctly. Redefinitions can restore continuity. To check continuity at any point a, use this checklist: ensure f(a) is defined, compute $\lim_{x→a}$ f(x) and confirm it exists, then check if $\lim_{x→a}$ f(x) = f(a).
8. For $o(x)=\begin{cases}x^2-1,&x<0\\-1,&x=0\\1-x^2,&x>0\end{cases}$, is $o$ continuous at $x=0$, and why?
   Answer: No; $\lim_{x\to0^-}o(x)=-1$ and $\lim_{x\to0^+}o(x)=1$, so the limit does not exist. Continuity at a point: f(a) defined, limit exists, equal. For o(x) at x = 0, o(0) = -1, but left x²-1 → -1, right 1-x² → 1, DNE, discontinuous. Common omission: not verifying one-sided limits match. Parabolic pieces disagree. Symmetry matters for limit existence. To check continuity at any point a, use this checklist: ensure f(a) is defined, compute $\lim_{x→a}$ f(x) and confirm it exists, then check if $\lim_{x→a}$ f(x) = f(a).
9. For $Y(x)=\frac{1}{x-1}$, is $Y$ continuous at $x=1$, and why?
   Answer: No; $Y(1)$ is undefined, so it cannot be continuous at $1$. Continuity requires the function to be defined at the point, the limit to exist, and equality between them. For Y(x) at x = 1, Y(1) is undefined due to division by zero, so it cannot be continuous there, even though the limit diverges. A frequent omission is focusing only on the limit without confirming the function is defined at the point. Rational functions have discontinuities where the denominator is zero. This highlights why domain restrictions affect continuity. To check continuity at any point a, use this checklist: ensure f(a) is defined, compute $\lim_{x→a}$ f(x) and confirm it exists, then check if $\lim_{x→a}$ f(x) = f(a).
10. Let $g(x)=\begin{cases}x+2,&x<1\\4,&x=1\\2x,&x>1\end{cases}$. Is $g$ continuous at $x=1$, and why?​​​
    Answer: No; $\lim_{x\to1^-}g(x)=3$ and $\lim_{x\to1^+}g(x)=2$, so $\lim_{x\to1}g(x)$ does not exist. To determine continuity at x = 1, we must check if g(1) is defined, if lim[x→1] g(x) exists, and if they are equal. From the piecewise function, g(1) = 4, so the function is defined at x = 1. For the limit to exist, the left-hand and right-hand limits must be equal: lim[x→1⁻] g(x) = lim[x→1⁻] (x + 2) = 3 and lim[x→1⁺] g(x) = lim[x→1⁺] (2x) = 2. Since 3 ≠ 2, the two-sided limit does not exist, failing condition 2 for continuity. Students often assume piecewise functions are continuous without checking one-sided limits at transition points. Always verify continuity using: (1) Check f(a) exists, (2) Calculate both one-sided limits, (3) Ensure all three values match.
11. For $g(x)=\begin{cases}x^2+1,&x<1\\5,&x=1\\x+1,&x>1\end{cases}$, is $g$ continuous at $x=1$, and why?
    Answer: No; $\lim_{x\to1^-}g(x)=2$ and $\lim_{x\to1^+}g(x)=2$, but $g(1)=5$. Continuity requires all three: defined f(a), existing limit, and equality. For g(x) at x = 1, g(1) = 5, but both sides limit to 2 ≠ 5, so discontinuous despite limit existing. A frequent oversight is not comparing the limit to f(a) after confirming the limit. Different pieces approach the same but not matching the point. This shows removable discontinuity if redefined. To check continuity at any point a, use this checklist: ensure f(a) is defined, compute $\lim_{x→a}$ f(x) and confirm it exists, then check if $\lim_{x→a}$ f(x) = f(a).
12. Let $f(x)=\begin{cases}\sin x/x,&x\ne0\\1,&x=0\end{cases}$. Is $f$ continuous at $x=0$, and why?
    Answer: Yes; $f(0)=1$ and $\lim_{x\to0}f(x)=1$, so $\lim_{x\to0}f(x)=f(0)$. Continuity at x=0 requires f(0) defined, limit existing, and equality. Here, f(0)=1, and the known limit of sin(x)/x is 1, so all conditions hold, making f continuous. The piecewise definition fills the hole at 0 where sin(x)/x is undefined. A frequent omission is assuming the function is undefined at 0 without checking the piecewise rule. This extends the sinc function continuously. Use this transferable checklist: (1) Check if f(a) is defined, (2) Confirm the limit exists, (3) Verify limit equals f(a).
13. Let $d(x)=\begin{cases}\frac{x^2-25}{x-5},&x\ne5\\10,&x=5\end{cases}$. Is $d$ continuous at $x=5$, and why?
    Answer: Yes; $\lim_{x\to5}d(x)=10$ and $d(5)=10$. For continuity, verify f(a) defined, limit exists, and equals f(a). At x = 5 for d(x), d(5) = 10, and simplified limit (x+5) → 10, so continuous. Students often forget to simplify rational expressions before taking limits. The removable discontinuity is fixed by redefining. This shows how holes can be filled for continuity. To check continuity at any point a, use this checklist: ensure f(a) is defined, compute $\lim_{x→a}$ f(x) and confirm it exists, then check if $\lim_{x→a}$ f(x) = f(a).
14. Let $s(x)=\begin{cases}2x+1,&x\le-1\\x^2,&x>-1\end{cases}$. Is $s$ continuous at $x=-1$, and why?
    Answer: No; $\lim_{x\to-1^-}s(x)=-1$ and $\lim_{x\to-1^+}s(x)=1$, so the limit does not exist. For continuity at x = -1, we check if s(-1) is defined, if lim[x→-1] s(x) exists, and if they equal. Since -1 satisfies x ≤ -1, we use the first formula: s(-1) = 2(-1) + 1 = -1, so the function is defined. For one-sided limits: lim[x→-1⁻] s(x) = lim[x→-1⁻] (2x + 1) = -1 and lim[x→-1⁺] s(x) = lim[x→-1⁺] x² = 1. Since -1 ≠ 1, the two-sided limit doesn't exist, violating condition 2. The common error is assuming piecewise functions are continuous at transition points without checking both sides. Use the checklist: (1) s(-1) = -1 ✓, (2) Left limit = -1, right limit = 1, so limit DNE ✗, (3) N/A.
15. For $U(x)=\begin{cases}x^2+2,&x\ne-3\\11,&x=-3\end{cases}$, is $U$ continuous at $x=-3$, and why?
    Answer: Yes; $\lim_{x\to-3}U(x)=11$ and $U(-3)=11$. Continuity at a point requires that the function is defined at that point, the limit exists as x approaches that point, and the limit equals the function's value there. For U(x) at x = -3, U(-3) = 11 is defined, and the limit is lim (x² + 2) = 11, which matches U(-3), so it is continuous. A common omission is forgetting to check if the limit equals the redefined value at the point, as the piecewise definition might suggest a hole otherwise. Here, since the limit of the quadratic equals the given value, continuity holds. Remember, redefining a function at a single point can make it continuous if it fills the hole correctly. To verify continuity at any point a, use this checklist: ensure f(a) is defined, compute $\lim_{x→a}$ f(x) and confirm it exists, then check if $\lim_{x→a}$ f(x) = f(a).
16. Let $V(x)=\begin{cases}0,&x\le0\\frac{1}{x},&x>0\end{cases}$. Is $V$ continuous at $x=0$, and why?
    Answer: No; $\lim_{x\to0^+}V(x)$ does not exist (diverges), so $\lim_{x\to0}V(x)$ does not exist. A function is continuous at x = a if f(a) exists, the limit as x approaches a exists, and they are equal. For V(x) at x = 0, V(0) = 0 is defined, but the right-hand limit diverges to infinity while the left-hand limit is 0, so the overall limit does not exist, making it discontinuous. Students often omit checking both one-sided limits for existence, assuming a defined value suffices. Here, the divergence from the right prevents the limit from existing. Continuity fails when limits behave differently from each side, like in step functions or asymptotes. To check continuity at any point a, use this checklist: ensure f(a) is defined, compute $\lim_{x→a}$ f(x) and confirm it exists, then check if $\lim_{x→a}$ f(x) = f(a).
17. Let $r(x)=\begin{cases}\dfrac{1}{x-3},&x\ne3\\0,&x=3\end{cases}$. Is $r$ continuous at $x=3$, and why?​​
    Answer: No; $\lim_{x\to3}r(x)$ does not exist (it diverges), so it cannot equal $r(3)=0$. To determine continuity at x = 3, we check: (1) r(3) exists, (2) lim[x→3] r(x) exists, and (3) they are equal. From the definition, r(3) = 0, satisfying condition (1). For the limit, we examine lim[x→3] 1/(x - 3): as x approaches 3 from the right, 1/(x - 3) → +∞, and from the left, 1/(x - 3) → -∞. Since the one-sided limits are infinite and opposite in sign, lim[x→3] r(x) does not exist, violating condition (2). When the limit doesn't exist, it cannot equal r(3), so r is not continuous at 3. Students sometimes think assigning a value at a vertical asymptote creates continuity, but the limit must exist finitely. Continuity checklist: ✓ Function defined at point, ✗ Limit exists, ✗ Limit equals function value.
18. For $t(x)=\sqrt{x-1}$ (domain $x\ge1$), is $t$ continuous at $x=1$, and why?
    Answer: Yes; $t(1)=0$ and $\lim_{x\to1^+}t(x)=0$, so the limit equals the value on its domain. Continuity at endpoint x=1 for domain x≥1 requires f(1) defined, the one-sided limit from the right exists, and they equal. t(1)=0, and right limit is 0, so it is continuous (using one-sided definition for endpoints). No left limit exists due to domain, but that's acceptable. A common omission is insisting on two-sided limits even at boundaries. This square root function is continuous on its domain. Checklist: (1) f(a) defined? (2) Appropriate limit exists? (3) Equals f(a)?
19. Let $f(x)=\begin{cases}\frac{x^2-9}{x-3},&x\ne3\\5,&x=3\end{cases}$. Is $f$ continuous at $x=3$, and why?
    Answer: No; $\lim_{x\to3}f(x)=6$ exists, but $f(3)=5$, so $\lim_{x\to3}f(x)\ne f(3)$. To check continuity at x = 3, we need to verify three conditions: (1) f(3) is defined, (2) lim[x→3] f(x) exists, and (3) lim[x→3] f(x) = f(3). From the piecewise definition, f(3) = 5, so condition 1 is satisfied. For the limit, we evaluate lim[x→3] (x² - 9)/(x - 3) = lim[x→3] (x + 3)(x - 3)/(x - 3) = lim[x→3] (x + 3) = 6, so condition 2 is satisfied. However, since lim[x→3] f(x) = 6 but f(3) = 5, we have 6 ≠ 5, violating condition 3. Many students forget to check whether the limit equals the function value, focusing only on whether the limit exists. Remember the continuity checklist: (1) Is f(a) defined? (2) Does lim[x→a] f(x) exist? (3) Does lim[x→a] f(x) = f(a)?
20. Let $k(x)=\begin{cases}x^2,&x<2\\4,&x=2\\5,&x>2\end{cases}$. Is $k$ continuous at $x=2$, and why?
    Answer: No; $\lim_{x\to2^-}k(x)=4$ and $\lim_{x\to2^+}k(x)=5$, so the limit does not exist. For continuity, the limit must exist and equal f(a), with f(a) defined. At x = 2 for k(x), k(2) = 4, but left limit 4 and right 5 differ, so limit DNE, discontinuous. Often, students forget to check one-sided agreement in piecewise constants. This is a jump discontinuity. Matching pieces are crucial. To check continuity at any point a, use this checklist: ensure f(a) is defined, compute $\lim_{x→a}$ f(x) and confirm it exists, then check if $\lim_{x→a}$ f(x) = f(a).

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