Confirming Continuity over an Interval
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AP Calculus AB › Confirming Continuity over an Interval
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?
No; $\lim_{x\to1^-}Q(x)=0$ and $\lim_{x\to1^+}Q(x)=1$, so the limit does not exist.
Yes; $\lim_{x\to1}Q(x)=0$ exists and equals $Q(1)=0$.
No; $\lim_{x\to1}Q(x)=-1$ but $Q(1)=0$.
No; $\lim_{x\to1}Q(x)=1$ but $Q(1)=0$.
No; $Q(1)$ is undefined.
Explanation
Q at 1: Q(1)=0, both limits 0 matches. Continuous. Omission: assuming symmetry causes issue. Checklist: f(a), sides, match.
Let $V(x)=\begin{cases}x^2+4,&x\le0\\4-x,&x>0\end{cases}$. Is $V$ continuous at $x=0$, and why?
No; $\lim_{x\to0}V(x)=0$ but $V(0)=4$.
No; $\lim_{x\to0^-}V(x)=4$ and $\lim_{x\to0^+}V(x)=3$, so the limit does not exist.
No; $V(0)$ is undefined.
No; $\lim_{x\to0}V(x)=5$ but $V(0)=4$.
Yes; $\lim_{x\to0}V(x)=4$ exists and equals $V(0)=4$.
Explanation
V at 0: V(0)=4, left 4, right 4 matches. Continuous. Omission: miscalculating right limit as 3. Checklist: f(a), sides, match.
Let $E(x)=\begin{cases}\frac{x^2-25}{x-5},&x\ne5\\10,&x=5\end{cases}$. Is $E$ continuous at $x=5$, and why?
No; $E(5)$ is undefined.
No; $\lim_{x\to5}E(x)$ does not exist because the denominator is zero.
Yes; $\lim_{x\to5}E(x)=10$ exists and equals $E(5)=10$.
No; $\lim_{x\to5}E(x)=0$ but $E(5)=10$.
No; $\lim_{x\to5}E(x)=10$ but $E(5)=5$.
Explanation
Continuity of E at x = 5 demands the limit exists, E(5) is defined, and they are equal. E(5) = 10 is defined. Simplifying $(x^2$ - 25)/(x - 5) = x + 5 for x ≠ 5, the limit as x approaches 5 is 10. Since the limit equals E(5), it is continuous. A common omission is failing to simplify removable discontinuities before evaluating the limit. Another is assuming undefined at the point means no continuity without checking the limit. Checklist: confirm f(a) defined, compute the limit (simplifying if needed), ensure limit equals f(a).
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?
No; $\lim_{x\to0}P(x)=4$ but $P(0)=-4$.
Yes; $\lim_{x\to0}P(x)=-4$ exists and equals $P(0)=-4$.
No; $\lim_{x\to0}P(x)$ does not exist because one-sided limits differ.
No; $P(0)$ is undefined.
No; $\lim_{x\to0}P(x)=0$ but $P(0)=-4$.
Explanation
P at 0: P(0)=-4, simplified limit x-4=-4 matches. Continuous. Omission: simplifying. Checklist: f(a), limit, equality.
Let $M(x)=\begin{cases}x^2,&x<1\\2,&x\ge1\end{cases}$. Is $M$ continuous at $x=1$, and why?
No; $\lim_{x\to1^-}M(x)=1$ and $\lim_{x\to1^+}M(x)=2$, so the limit does not exist.
No; $M(1)$ is undefined.
Yes; $\lim_{x\to1}M(x)=2$ equals $M(1)=2$.
No; $\lim_{x\to1}M(x)=1$ but $M(1)=2$.
No; $\lim_{x\to1}M(x)=3$ but $M(1)=2$.
Explanation
M at 1: M(1)=2, left limit 1, right 2, no limit. Not continuous. Omission: boundary inclusion. Checklist: f(a), sides agree, match.
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?
No; $\lim_{x\to2}J(x)=5$ but $J(2)=4$.
No; $\lim_{x\to2^-}J(x)=5$ and $\lim_{x\to2^+}J(x)=4$, so the limit does not exist.
Yes; $\lim_{x\to2}J(x)=4$ exists and equals $J(2)=4$.
No; $\lim_{x\to2}J(x)=3$ but $J(2)=4$.
No; $J(2)$ is undefined.
Explanation
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 $X(x)=\begin{cases}2x+3,&x< -1\\1,&x=-1\\2x+3,&x>-1\end{cases}$. Is $X$ continuous at $x=-1$, and why?
No; $\lim_{x\to-1}X(x)$ does not exist because one-sided limits differ.
No; $\lim_{x\to-1}X(x)=1$ exists but $X(-1)=5$.
Yes; $\lim_{x\to-1}X(x)=1$ exists and equals $X(-1)=1$.
No; $\lim_{x\to-1}X(x)=0$ but $X(-1)=1$.
No; $\lim_{x\to-1}X(x)=1$ exists but $X(-1)=1$ is undefined.
Explanation
X at -1: X(-1)=1, limit 2x+3=1 matches. Continuous. Omission: thinking middle differs. Checklist: f(a), limit, equality.
For $s(x)=\begin{cases}\sin x,&x\ne0\\1,&x=0\end{cases}$, is $s$ continuous at $x=0$, and why?
No; $\lim_{x\to0}s(x)$ does not exist because $\sin x$ oscillates.
Yes; $\lim_{x\to0}s(x)=1=s(0)$.
Yes; $\sin x$ is continuous, so redefining $s(0)$ cannot affect continuity.
No; $s(0)$ is undefined.
No; $\lim_{x\to0}s(x)=0\ne s(0)=1$.
Explanation
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?
For $r(x)=\frac{x^2-1}{x-1}$ when $x\ne1$ and $r(1)=0$, is $r$ continuous at $x=1$, and why?
Yes; $r(1)$ exists, so $r$ is continuous at $1$.
No; $\lim_{x\to1}r(x)=2$ but $r(1)=0$.
No; $\lim_{x\to1}r(x)$ does not exist because the denominator is $0$ at $x=1$.
No; $r(1)$ is undefined.
Yes; $\lim_{x\to1}r(x)=0$ equals $r(1)$.
Explanation
For r(x) = (x² - 1)/(x - 1) when x ≠ 1, we can factor and simplify: r(x) = (x + 1)(x - 1)/(x - 1) = x + 1 for x ≠ 1. Therefore, lim[x→1] r(x) = lim[x→1] (x + 1) = 2. We're told r(1) = 0, which is defined. However, since the limit (2) doesn't equal r(1) = 0, the function is not continuous at x = 1. This is a removable discontinuity that hasn't been properly removed—to make r continuous, we would need to define r(1) = 2, not 0. Students often assume that defining the function at a point automatically makes it continuous there, but the definition must match the limit. Continuity checklist: ✓ r(1) defined, ✓ Limit exists (= 2), ✗ Limit ≠ r(1) (2 ≠ 0).
Let $s(x)=\begin{cases}2x,&x\le1\\2,&x>1\end{cases}$. Is $s$ continuous at $x=1$, and why?
Yes; $\lim_{x\to1}s(x)$ exists and equals $2=s(1)$.
No; $\lim_{x\to1^-}s(x)=2$ and $\lim_{x\to1^+}s(x)=2$, but $s(1)=1$.
No; $\lim_{x\to1}s(x)$ does not exist because the rule changes at $x=1$.
No; $s(1)$ is undefined.
Yes; $s$ is continuous at $1$ because both pieces are linear.
Explanation
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).