Confirming Continuity over an Interval - AP Calculus AB

Define $t(x)=\begin{cases}(x-4)^2,&x\ne4\\7,&x=4\end{cases}$ . Is $t$ continuous at $x=4$ , and why?
Answer: No; $\lim_{x\to4}t(x)=0$ but $t(4)=7$ . We have t(x) = (x - 4)² for x ≠ 4 and t(4) = 7. To find the limit as x approaches 4, we evaluate lim[x→4] (x - 4)² = (4 - 4)² = 0. The function is defined at x = 4 with t(4) = 7. However, the limit (0) doesn't equal t(4) = 7, so t is not continuous at x = 4. This is another removable discontinuity where the function has been defined incorrectly at the point—to be continuous, we would need t(4) = 0. Students might think that since (x - 4)² is a continuous polynomial, defining t(4) = 7 maintains continuity, but the definition must match what the function naturally approaches. Continuity checklist: ✓ t(4) defined, ✓ Limit exists (= 0), ✗ Limit ≠ t(4) (0 ≠ 7).
Let $N(x)=\begin{cases}e^x,&x\ne0\\1,&x=0\end{cases}$ . Is $N$ continuous at $x=0$ , and why?
Answer: Yes; $\lim_{x\to0}N(x)=1$ exists and equals $N(0)=1$ . N at 0: N(0)=1, limit $e^0$ =1 matches. Continuous. Omission: unnecessary piecewise check. Checklist: f(a), limit, equality.
Let $J(x)=\begin{cases}x^2+1,&x<2\\4,&x=2\\x^2,&x>2\end{cases}$ . Is $J$ continuous at $x=2$ , and why?
Answer: No; $\lim_{x\to2^-}J(x)=5$ and $\lim_{x\to2^+}J(x)=4$ , so the limit does not exist. J at 2: J(2)=4 defined, left limit 4+1=5, right 4, differ so no limit. Not continuous. Omission: not checking sides. Checklist: f(a), left/right agree, match.
Let $T(x)=\begin{cases}3x,&x<2\\6,&x=2\\3x+1,&x>2\end{cases}$ . Is $T$ continuous at $x=2$ , and why?
Answer: No; $\lim_{x\to2^-}T(x)=6$ and $\lim_{x\to2^+}T(x)=7$ , so the limit does not exist. T at 2: T(2)=6, left 6, right 7, no limit. Not continuous. Omission: close values. Checklist: f(a), sides agree, match.
Let $D(x)=\begin{cases}x^2-1,&x<1\\0,&x=1\\2-x,&x>1\end{cases}$ . Is $D$ continuous at $x=1$ , and why?
Answer: No; $\lim_{x\to1^-}D(x)=0$ and $\lim_{x\to1^+}D(x)=1$ , so the limit does not exist. For D to be continuous at x = 1, the limit must exist, D(1) must be defined, and they must match. D(1) = 0 is defined, but the left limit is $1^2$ - 1 = 0 and the right limit is 2 - 1 = 1, so the limit does not exist due to differing one-sided limits. Therefore, D is not continuous at x = 1. A common omission is not checking both one-sided limits separately when the function is piecewise. People often assume the function value dictates continuity without limit agreement. Use this checklist for continuity: verify f(a) exists, ensure left and right limits exist and equal each other, and confirm that value matches f(a).
For $f(x)=\frac{x^2-9}{x-3}$ when $x\ne3$ and $f(3)=6$ , is $f$ continuous at $x=3$ , and why?
Answer: Yes; $\lim_{x\to3}f(x)$ exists and equals $f(3)=6$ . 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). We're given that f(3) = 6, so condition 1 is met. For the limit, we can factor the numerator: f(x) = (x² - 9)/(x - 3) = (x + 3)(x - 3)/(x - 3) = x + 3 for x ≠ 3. Therefore, lim[x→3] f(x) = lim[x→3] (x + 3) = 6. Since the limit equals f(3) = 6, all three conditions are satisfied. A common mistake is assuming the function is discontinuous just because the original expression has x - 3 in the denominator, but the removable discontinuity has been properly addressed by defining f(3) = 6. Continuity checklist: ✓ Function defined at the point, ✓ Limit exists, ✓ Limit equals function value.
Let $Q(x)=\begin{cases}1-x,&x<1\\0,&x=1\\x-1,&x>1\end{cases}$ . Is $Q$ continuous at $x=1$ , and why?
Answer: Yes; $\lim_{x\to1}Q(x)=0$ exists and equals $Q(1)=0$ . Q at 1: Q(1)=0, both limits 0 matches. Continuous. Omission: assuming symmetry causes issue. Checklist: f(a), sides, match.
Let $C(x)=\begin{cases}\frac{x}{x^2},&x\ne0\\0,&x=0\end{cases}$ . Is $C$ continuous at $x=0$ , and why?
Answer: No; $\lim_{x\to0}C(x)$ does not exist (it is unbounded) though $C(0)=0$ . Continuity at x = 0 for C requires the limit to exist, C(0) to be defined, and them to be equal. C(0) = 0 is defined, but the limit of 1/x as x approaches 0 does not exist because it approaches +∞ from the right and -∞ from the left, making it unbounded. Thus, the function is not continuous despite the defined value. A common omission is ignoring that the limit must be finite and exist for continuity, mistaking a defined point for filling a 'hole'. Explanations often overlook checking one-sided limits for potential divergences. To verify continuity at a point, always: ensure f(a) is defined, confirm the limit exists (left and right agree on a finite value), and check if limit equals f(a).
Let $R(x)=\begin{cases}\sqrt{4-x},&x<4\\0,&x=4\\x-4,&x>4\end{cases}$ . Is $R$ continuous at $x=4$ , and why?
Answer: Yes; $\lim_{x\to4}R(x)=0$ exists and equals $R(4)=0$ . R at 4: R(4)=0, left √0=0, right 0 matches. Continuous. Omission: domain for sqrt. Checklist: f(a), limits, equality.
Let $Y(x)=\begin{cases}\frac{|x|}{2},&x\ne0\\1,&x=0\end{cases}$ . Is $Y$ continuous at $x=0$ , and why?
Answer: No; $\lim_{x\to0}Y(x)=0$ exists but $Y(0)=1$ . Y at 0: limit |x|/2=0, Y(0)=1 ≠0, not continuous. Omission: thinking limit=1. Checklist: f(a), limit, equality.
Let $V(x)=\begin{cases}x^2+4,&x\le0\\4-x,&x>0\end{cases}$ . Is $V$ continuous at $x=0$ , and why?
Answer: Yes; $\lim_{x\to0}V(x)=4$ exists and equals $V(0)=4$ . V at 0: V(0)=4, left 4, right 4 matches. Continuous. Omission: miscalculating right limit as 3. Checklist: f(a), sides, match.
Let $g(x)=\begin{cases}x+1,&x<2\\5,&x=2\\x^2+1,&x>2\end{cases}$ . Is $g$ continuous at $x=2$ , and why?
Answer: No; $\lim_{x\to2^-}g(x)=3$ and $\lim_{x\to2^+}g(x)=5$ , so the limit does not exist. For piecewise functions, continuity at a transition point requires checking both one-sided limits and ensuring they equal the function value. At x = 2, we have g(2) = 5 from the middle piece. From the left: lim[x→2⁻] g(x) = lim[x→2⁻] (x + 1) = 3. From the right: lim[x→2⁺] g(x) = lim[x→2⁺] (x² + 1) = 5. Since the left-hand limit (3) doesn't equal the right-hand limit (5), the two-sided limit doesn't exist. Without a limit, continuity fails even though g(2) is defined. Many students incorrectly think that having g(2) defined is sufficient for continuity. Continuity checklist for piecewise functions: ✓ Function defined at transition, ✗ Left limit = Right limit (failed here), ✗ Common limit = Function value.
Let $s(x)=\begin{cases}2x,&x\le1\\2,&x>1\end{cases}$ . Is $s$ continuous at $x=1$ , and why?
Answer: Yes; $\lim_{x\to1}s(x)$ exists and equals $2=s(1)$ . For this piecewise function, we need to check continuity at the transition point x = 1. From the definition, s(1) = 2(1) = 2 using the first piece (x ≤ 1). The left-hand limit: lim[x→1⁻] s(x) = lim[x→1⁻] 2x = 2. The right-hand limit: lim[x→1⁺] s(x) = lim[x→1⁺] 2 = 2. Since both one-sided limits equal 2, the two-sided limit exists and equals 2. With s(1) = 2 matching this limit, all continuity conditions are satisfied. The function smoothly transitions from the line y = 2x to the constant y = 2 at x = 1. Students sometimes incorrectly assume piecewise functions are automatically discontinuous at transition points. Continuity checklist for piecewise: ✓ s(1) defined (= 2), ✓ Left and right limits equal (both 2), ✓ Common limit equals s(1).
For $s(x)=\begin{cases}\sin x,&x\ne0\\1,&x=0\end{cases}$ , is $s$ continuous at $x=0$ , and why?
Answer: No; $\lim_{x\to0}s(x)=0\ne s(0)=1$ . To determine continuity at x = 0, we verify: (1) s(0) exists, (2) lim[x→0] s(x) exists, and (3) they are equal. The function defines s(0) = 1, so condition (1) is met. For the limit, since s(x) = sin x for x ≠ 0, we have lim[x→0] s(x) = lim[x→0] sin x = sin(0) = 0. We have s(0) = 1 but lim[x→0] s(x) = 0, so 1 ≠ 0, violating condition (3). This is a classic example where redefining a continuous function at a single point creates a discontinuity—the natural value of sin(0) = 0 has been changed to 1. Remember the continuity checklist: (1) Function value exists? (2) Limit exists? (3) Are they equal?
Let $K(x)=\begin{cases}x+2,&x\ne-3\\0,&x=-3\end{cases}$ . Is $K$ continuous at $x=-3$ , and why?
Answer: No; $\lim_{x\to-3}K(x)=-1$ exists but $K(-3)=0$ . K at -3: K(-3)=0 defined, limit of x+2 = -1 ≠0. Not continuous. Omission: assuming redefinition matches limit. Checklist: f(a), limit, equality.
For $u(x)=\begin{cases}x^3,&x\le2\\8+(x-2),&x>2\end{cases}$ , is $u$ continuous at $x=2$ , and why?
Answer: Yes; $u(2)=8$ and $\lim_{x\to2^-}u(x)=\lim_{x\to2^+}u(x)=8$ . To check continuity at x = 2, we verify: (1) u(2) exists, (2) lim[x→2] u(x) exists, and (3) they are equal. Since 2 falls in the x ≤ 2 case, u(2) = 2³ = 8, satisfying condition (1). For the limit, we compute: lim[x→2⁻] u(x) = lim[x→2⁻] x³ = 2³ = 8, and lim[x→2⁺] u(x) = lim[x→2⁺] [8 + (x - 2)] = 8 + (2 - 2) = 8. Both one-sided limits equal 8, so the two-sided limit exists and equals 8. With u(2) = 8 and lim[x→2] u(x) = 8, all conditions are satisfied, making u continuous at x = 2. The key insight is that the second piece was carefully designed to "pick up" where the first piece left off. Apply the continuity checklist: (1) Function value exists? (2) Limit exists? (3) Are they equal?
Let $H(x)=\begin{cases}2x+1,&x\le3\\7,&x>3\end{cases}$ . Is $H$ continuous at $x=3$ , and why?
Answer: Yes; $\lim_{x\to3}H(x)=7$ exists and equals $H(3)=7$ . Continuity at x = 3 for H requires limit existence, defined H(3), and equality. H(3) = 2*3 + 1 = 7. Left limit 2*3 + 1 = 7, right limit 7, so limit = 7 matches. Continuous. Common omission: misapplying the piece at the boundary. Checklist: compute f(a) using correct piece, verify limit via one-sides, confirm equality.
Let $P(x)=\begin{cases}\frac{x^2-4x}{x},&x\ne0\\-4,&x=0\end{cases}$ . Is $P$ continuous at $x=0$ , and why?
Answer: Yes; $\lim_{x\to0}P(x)=-4$ exists and equals $P(0)=-4$ . P at 0: P(0)=-4, simplified limit x-4=-4 matches. Continuous. Omission: simplifying. Checklist: f(a), limit, equality.
For $u(x)=\frac{1}{x-5}$ , is $u$ continuous at $x=5$ , and why?
Answer: No; $u(5)$ is undefined, so continuity at $5$ fails. The function u(x) = 1/(x - 5) is a rational function with a vertical asymptote at x = 5. At x = 5, the denominator equals zero, making u(5) undefined—we cannot evaluate 1/0. Since the function isn't even defined at x = 5, it cannot be continuous there. Additionally, as x approaches 5, the function values grow without bound (approaching ±∞ depending on direction), so lim[x→5] u(x) doesn't exist as a finite value. This is a non-removable discontinuity (vertical asymptote) that cannot be fixed by defining u(5). Students sometimes confuse undefined expressions with limits that equal zero. Continuity checklist: ✗ u(5) undefined (first condition fails immediately), ✗ Limit doesn't exist (approaches ±∞), ✗ Cannot compare limit to function value.
Let $M(x)=\begin{cases}x^2,&x<1\\2,&x\ge1\end{cases}$ . Is $M$ continuous at $x=1$ , and why?
Answer: No; $\lim_{x\to1^-}M(x)=1$ and $\lim_{x\to1^+}M(x)=2$ , so the limit does not exist. M at 1: M(1)=2, left limit 1, right 2, no limit. Not continuous. Omission: boundary inclusion. Checklist: f(a), sides agree, match.

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