Defining Continuity at a Point
Help Questions
AP Calculus AB › Defining Continuity at a Point
Let $f(x)=\begin{cases}\sin x/x,&x\ne0\\1,&x=0\end{cases}$. Is $f$ continuous at $x=0$, and why?
No; $\lim_{x\to0}f(x)=0$ but $f(0)=1$.
No; $f(0)$ is undefined since $\sin 0/0$ is undefined.
Yes; the limit exists, so continuity holds even if $f(0)$ were different.
No; $\lim_{x\to0}\sin x/x$ does not exist because $\sin 0=0$.
Yes; $f(0)=1$ and $\lim_{x\to0}f(x)=1$, so $\lim_{x\to0}f(x)=f(0)$.
Explanation
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).
Let $J(x)=\begin{cases}3,&x<5\\3,&x=5\\3,&x>5\end{cases}$. Is $J$ continuous at $x=5$, and why?
No; the limit exists but $J(5)$ is undefined.
Yes; $\lim_{x\to5}J(x)=3$ and $J(5)=3$.
No; $\lim_{x\to5}J(x)=5$ but $J(5)=3$.
Yes; $J(5)=3$ so continuity does not require a limit.
No; $\lim_{x\to5}J(x)$ does not exist because the function is constant.
Explanation
Continuity at x=5: J(5)=3, limit 3 from constants, matches, continuous. Essentially constant function. Omission: overcomplicating pieces. Continuous everywhere. Strategy: (1) f(a)? (2) Limit? (3) Equal?
For $S(x)=\begin{cases}\frac{x^2-16}{x-4},&x\ne4\\8,&x=4\end{cases}$, is $S$ continuous at $x=4$, and why?
No; $\lim_{x\to4}S(x)$ does not exist because the denominator is $0$.
Yes; $\lim_{x\to4}S(x)=8$ and $S(4)=8$.
No; $\lim_{x\to4}S(x)=4$ but $S(4)=8$.
Yes; $S(4)=8$ so the limit must be $8$.
No; $S(4)$ is undefined since the fraction is undefined at $4$.
Explanation
At x=4, S(4)=8, limit 8 from x+4, continuous. Fills hole. Common: forgetting simplification. Removable fixed. Checklist: (1) Defined? (2) Limit? (3) Equals?
Let $v(x)=\begin{cases}\sqrt{x-1},&x\ge1\\0,&x<1\end{cases}$. Is $v$ continuous at $x=1$, and why?
No; $\lim_{x\to1}v(x)$ does not exist because $\sqrt{x-1}$ is not defined for $x<1$.
Yes; $\lim_{x\to1}v(x)=0$ but $v(1)=1$, so it is continuous.
Yes; $\lim_{x\to1^-}v(x)=0$, $\lim_{x\to1^+}v(x)=0$, and $v(1)=0$, so the limit exists and equals the value.
No; $\lim_{x\to1}v(x)=1$, but $v(1)=0$.
No; $v(1)$ is undefined because square roots cannot include $0$.
Explanation
To determine continuity at x = 1, we check: (1) v(1) exists, (2) lim[x→1] v(x) exists, and (3) they are equal. Since 1 is included in the first piece (x ≥ 1), we have v(1) = √(1-1) = √0 = 0, satisfying condition (1). For the limit: from the right, lim[x→1⁺] v(x) = lim[x→1⁺] √(x-1) = √0 = 0, and from the left, lim[x→1⁻] v(x) = lim[x→1⁻] 0 = 0. Both one-sided limits equal 0, so lim[x→1] v(x) = 0. Since v(1) = 0 and the limit = 0, all conditions are met—v is continuous at x = 1. The function cleverly avoids issues with √(x-1) for x < 1 by defining it as 0 there. Continuity checklist: ✓ Function defined at point, ✓ Limit exists, ✓ Limit equals function value.
Let $p(x)=\begin{cases}\dfrac{\sin x}{x},&x\ne0\\1,&x=0\end{cases}$. Is $p$ continuous at $x=0$, and why?
Yes; $\lim_{x\to0}p(x)=0$ and $p(0)=0$.
Yes; $\lim_{x\to0}p(x)=1$ and $p(0)=1$, so the limit exists and equals the value.
No; $p(0)$ is undefined because $\sin 0/0$ is undefined.
No; $\lim_{x\to0}p(x)$ does not exist because $\sin x$ oscillates.
No; $\lim_{x\to0}p(x)=0$ but $p(0)=1$.
Explanation
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.
Let $f(x)=\begin{cases}x^2-1,&x\ne1\\3,&x=1\end{cases}$. Is $f$ continuous at $x=1$, and why?
No; $f(1)$ is undefined, so continuity fails.
Yes; $\lim_{x\to1}f(x)=3$ and $f(1)=3$.
Yes; $f(1)$ exists and equals $0$, so $f$ is continuous.
No; $\lim_{x\to1}f(x)$ does not exist because $x^2-1$ is a polynomial.
No; $\lim_{x\to1}f(x)=0$ but $f(1)=3$, so the limit and value are not equal.
Explanation
To check continuity at x = 1, we need three conditions: (1) f(1) exists, (2) lim[x→1] f(x) exists, and (3) lim[x→1] f(x) = f(1). From the definition, f(1) = 3, so condition (1) is met. For the limit, we use the formula for x ≠ 1: lim[x→1] f(x) = lim[x→1] (x² - 1) = 1² - 1 = 0. Since the limit equals 0 but f(1) = 3, condition (3) fails—the limit and function value are not equal. A common mistake is assuming continuity just because f(1) is defined, but all three conditions must be satisfied. Continuity checklist: ✓ Function defined at point, ✓ Limit exists, ✗ Limit equals function value.
Let $t(x)=\begin{cases}x+2,&x\le -1\\x^2,&x>-1\end{cases}$. Is $t$ continuous at $x=-1$, and why?
Yes; $\lim_{x\to-1}t(x)=1$ and $t(-1)=1$, so the limit exists and equals the value.
No; $\lim_{x\to-1^-}t(x)=1$ and $\lim_{x\to-1^+}t(x)=1$, but $t(-1)=0$.
Yes; both pieces are continuous, so $t$ is continuous at $-1$.
No; $t(-1)$ is undefined because the definition switches at $-1$.
No; $\lim_{x\to-1}t(x)$ does not exist because one side is linear and the other is quadratic.
Explanation
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.
A function $u$ satisfies $\lim_{x\to5}u(x)=7$, but $u(5)$ is not defined. Is $u$ continuous at $5$, and why?
No; $u(5)$ does not exist, so continuity fails even though $\lim_{x\to5}u(x)$ exists.
No; the limit does not exist because $u(5)$ is undefined.
Yes; $u(5)$ must equal $7$ automatically since the limit is $7$.
Yes; the limit exists, so $u$ is continuous at $5$.
Yes; define $u(5)=0$ and it becomes continuous.
Explanation
For continuity at x = 5, we need three conditions: (1) u(5) exists, (2) lim[x→5] u(x) exists, and (3) lim[x→5] u(x) = u(5). We're told that lim[x→5] u(x) = 7, so condition (2) is satisfied. However, we're also told that u(5) is not defined, which means condition (1) fails immediately. Without u(5) existing, there's no value to compare with the limit, so condition (3) cannot be satisfied either. Therefore, u is not continuous at x = 5. A common mistake is thinking that if the limit exists, we can just define the function value to make it continuous—but the question asks about the given function, not a modified version. Continuity checklist: ✗ Function defined at point, ✓ Limit exists, ✗ Limit equals function value.
Let $r(x)=\begin{cases}\dfrac{1}{x-3},&x\ne3\\0,&x=3\end{cases}$. Is $r$ continuous at $x=3$, and why?
Yes; $r(3)=0$ and the limit is also $0$.
No; $\lim_{x\to3}r(x)=3$, but $r(3)=0$.
No; $\lim_{x\to3}r(x)$ does not exist (it diverges), so it cannot equal $r(3)=0$.
No; $r(3)$ is undefined because the denominator is $0$.
Yes; $\lim_{x\to3}r(x)$ exists because rational functions always have limits.
Explanation
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.
Let $r(x)=\begin{cases}\frac{1}{x-2},&x\ne2\\0,&x=2\end{cases}$. Is $r$ continuous at $x=2$, and why?
No; $r(2)$ is undefined because division by zero occurs in the formula.
No; $\lim_{x\to2}r(x)$ does not exist (it diverges), so continuity fails.
Yes; $r(2)=0$ and the limit equals $0$.
No; $\lim_{x\to2}r(x)=2$ exists, but $r(2)=0$.
Yes; $\lim_{x\to2}r(x)$ exists because rational functions are continuous.
Explanation
To check continuity at x = 2, we verify r(2) is defined, lim[x→2] r(x) exists, and they are equal. The function defines r(2) = 0, so condition 1 is satisfied. For the limit, we examine lim[x→2] 1/(x-2): as x approaches 2 from the right, x - 2 → 0⁺, so 1/(x-2) → +∞; as x approaches 2 from the left, x - 2 → 0⁻, so 1/(x-2) → -∞. Since the one-sided limits are different infinities, lim[x→2] r(x) does not exist, failing condition 2. Students sometimes think assigning a value at a discontinuity creates continuity, but the limit behavior cannot be "fixed" this way. Continuity requires: (1) f(a) defined ✓, (2) limit exists ✗ (diverges), (3) they match (N/A).