Confirming Continuity over an Interval
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AP Calculus AB › Confirming Continuity over an Interval
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 $N(x)=\begin{cases}e^x,&x\ne0\\1,&x=0\end{cases}$. Is $N$ continuous at $x=0$, and why?
Yes; $\lim_{x\to0}N(x)=1$ exists and equals $N(0)=1$.
No; $N(0)$ is undefined.
No; $\lim_{x\to0}N(x)=2$ but $N(0)=1$.
No; $\lim_{x\to0}N(x)$ does not exist because it oscillates.
No; $\lim_{x\to0}N(x)=0$ but $N(0)=1$.
Explanation
N at 0: N(0)=1, limit $e^0$=1 matches. Continuous. Omission: unnecessary piecewise check. Checklist: f(a), limit, equality.
Let $I(x)=\begin{cases}\frac{1}{x},&x\ne0\\1,&x=0\end{cases}$. Is $I$ continuous at $x=0$, and why?
No; $\lim_{x\to0}I(x)$ does not exist (unbounded), though $I(0)=1$.
Yes; $\lim_{x\to0}I(x)=1$ equals $I(0)=1$.
No; $I(0)$ is undefined.
No; $\lim_{x\to0}I(x)=-1$ but $I(0)=1$.
No; $\lim_{x\to0}I(x)=0$ but $I(0)=1$.
Explanation
I continuous at 0: limit exists, I(0) defined, equal. I(0)=1 defined, but 1/x limit unbounded, does not exist. Not continuous. Omission: assuming defined point fixes discontinuity. Checklist: f(a) defined, finite limit exists, equals f(a).
Let $F(x)=\begin{cases}\tan x,&x\ne\frac{\pi}{2}\\0,&x=\frac{\pi}{2}\end{cases}$. Is $F$ continuous at $x=\frac{\pi}{2}$, and why?
No; $\lim_{x\to\pi/2}F(x)=1$ but $F(\pi/2)=0$.
Yes; $\lim_{x\to\pi/2}F(x)=0$ equals $F(\pi/2)=0$.
No; $\lim_{x\to\pi/2}F(x)=-1$ but $F(\pi/2)=0$.
No; $F(\pi/2)$ is undefined.
No; $\lim_{x\to\pi/2}F(x)$ does not exist (vertical asymptote), though $F(\pi/2)=0$.
Explanation
To check if F is continuous at x = π/2, verify the limit exists, F(π/2) is defined, and they match. F(π/2) = 0 is defined, but tan x has a vertical asymptote at π/2, so the limit does not exist (approaches ±∞). Thus, F is not continuous. A common omission is not recognizing that infinite limits prevent existence for continuity. Explanations may forget to assess behavior near asymptotes. For continuity checks: ensure f(a) defined, verify limit exists finitely, confirm equality with f(a).
Let $K(x)=\begin{cases}x+2,&x\ne-3\\0,&x=-3\end{cases}$. Is $K$ continuous at $x=-3$, and why?
No; $\lim_{x\to-3}K(x)=1$ but $K(-3)=0$.
No; $\lim_{x\to-3}K(x)$ does not exist because one-sided limits differ.
No; $\lim_{x\to-3}K(x)=-1$ exists but $K(-3)=0$.
No; $K(-3)$ is undefined.
Yes; $\lim_{x\to-3}K(x)=0$ exists and equals $K(-3)=0$.
Explanation
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.
Let $Y(x)=\begin{cases}\frac{|x|}{2},&x\ne0\\1,&x=0\end{cases}$. Is $Y$ continuous at $x=0$, and why?
No; $Y(0)$ is undefined.
No; $\lim_{x\to0}Y(x)=2$ but $Y(0)=1$.
No; $\lim_{x\to0}Y(x)$ does not exist because of the absolute value.
No; $\lim_{x\to0}Y(x)=0$ exists but $Y(0)=1$.
Yes; $\lim_{x\to0}Y(x)=1$ equals $Y(0)=1$.
Explanation
Y at 0: limit |x|/2=0, Y(0)=1 ≠0, not continuous. Omission: thinking limit=1. Checklist: f(a), limit, equality.
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?
Yes; $\lim_{x\to2}T(x)=6$ exists and equals $T(2)=6$.
No; $\lim_{x\to2}T(x)=5$ but $T(2)=6$.
No; $\lim_{x\to2}T(x)=7$ but $T(2)=6$.
No; $\lim_{x\to2^-}T(x)=6$ and $\lim_{x\to2^+}T(x)=7$, so the limit does not exist.
No; $T(2)$ is undefined.
Explanation
T at 2: T(2)=6, left 6, right 7, no limit. Not continuous. Omission: close values. Checklist: f(a), sides agree, 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?
Yes; $\lim_{x\to5}E(x)=10$ exists and equals $E(5)=10$.
No; $\lim_{x\to5}E(x)$ does not exist because the denominator is zero.
No; $\lim_{x\to5}E(x)=10$ but $E(5)=5$.
No; $\lim_{x\to5}E(x)=0$ but $E(5)=10$.
No; $E(5)$ is undefined.
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 $G(x)=\begin{cases}x^2+3,&x< -2\\7,&x=-2\\x+5,&x>-2\end{cases}$. Is $G$ continuous at $x=-2$, and why?
Yes; $\lim_{x\to-2}G(x)=7$ exists and equals $G(-2)=7$.
No; $\lim_{x\to-2}G(x)=3$ but $G(-2)=7$.
No; $G(-2)$ is undefined.
No; $\lim_{x\to-2}G(x)=5$ but $G(-2)=7$.
No; $\lim_{x\to-2^-}G(x)=7$ and $\lim_{x\to-2^+}G(x)=3$, so the limit does not exist.
Explanation
For G continuous at x = -2, the limit must exist, G(-2) defined, and equal. G(-2) = 7 is defined, but left limit is $(-2)^2$ + 3 = 7 and right is -2 + 5 = 3, so limit does not exist. Hence, not continuous. Common omission: not computing one-sided limits for piecewise functions. Often, people check only the function value. Checklist: define f(a), check left/right limits agree, ensure match with f(a).
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; $V(0)$ is undefined.
Yes; $\lim_{x\to0}V(x)=4$ exists and equals $V(0)=4$.
No; $\lim_{x\to0}V(x)=0$ but $V(0)=4$.
No; $\lim_{x\to0}V(x)=5$ but $V(0)=4$.
No; $\lim_{x\to0^-}V(x)=4$ and $\lim_{x\to0^+}V(x)=3$, so the limit does not exist.
Explanation
V at 0: V(0)=4, left 4, right 4 matches. Continuous. Omission: miscalculating right limit as 3. Checklist: f(a), sides, match.