Defining Continuity at a Point
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AP Calculus AB › Defining Continuity at a Point
For $P(x)=\begin{cases}x-2,&x<2\\0,&x=2\\x-2,&x>2\end{cases}$, is $P$ continuous at $x=2$, and why?
No; $\lim_{x\to2}P(x)=2$ but $P(2)=0$.
Yes; $\lim_{x\to2}P(x)=0$ and $P(2)=0$.
No; $P(2)$ is undefined.
Yes; the limit exists, so $P$ is continuous even if $P(2)$ differed.
No; $\lim_{x\to2}P(x)$ does not exist because the function is linear.
Explanation
At x=2, P(2)=0, both limits 0 from x-2, continuous. Pieces agree at limit. Common: confusing with value. Continuous. Checklist: (1) Defined? (2) Limit? (3) Equals?
Let $w(x)=\begin{cases}x^3,&x<2\\8,&x\ge2\end{cases}$. Is $w$ continuous at $x=2$, and why?
No; $\lim_{x\to2^-}w(x)=8$ and $\lim_{x\to2^+}w(x)=0$, so the limit does not exist.
No; $\lim_{x\to2}w(x)=2$ but $w(2)=8$.
Yes; $w(2)=8$ so continuity is automatic.
Yes; $\lim_{x\to2^-}w(x)=8$, $\lim_{x\to2^+}w(x)=8$, and $w(2)=8$.
No; $\lim_{x\to2}w(x)=8$ but $w(2)$ is undefined.
Explanation
Continuity at x=2 needs w(2) defined, limit, and equality. w(2)=8, left limit from x³ is 8, right from constant 8 is 8, so continuous. The pieces connect smoothly. Omission: not checking right limit separately. Continuous here. Strategy: (1) f(a)? (2) Limit via sides? (3) Match?
For $e(x)=\begin{cases}\cos x,&x\ne0\\2,&x=0\end{cases}$, is $e$ continuous at $x=0$, and why?
No; $\lim_{x\to0}e(x)=1$ but $e(0)=2$.
Yes; cosine is continuous, so any value at $0$ makes it continuous.
Yes; $\lim_{x\to0}e(x)=2$ and $e(0)=2$.
No; $e(0)$ is undefined.
No; $\lim_{x\to0}e(x)$ does not exist because cosine oscillates.
Explanation
Continuity at a point means the limit matches the function value, with both defined and limit existing. For e(x) at x = 0, e(0) = 2, but lim cos x = 1 ≠ 2, so discontinuous. A common mistake is assuming trigonometric functions are continuous everywhere without checking redefinitions. Cosine is continuous, but the point change disrupts it. This illustrates point discontinuities in otherwise continuous functions. 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).
For $H(x)=\frac{|x|}{x}$ for $x\ne0$ and $H(0)=1$, is $H$ continuous at $x=0$, and why?
No; $\lim_{x\to0}H(x)=0$ but $H(0)=1$.
No; $\lim_{x\to0^-}H(x)=-1$ and $\lim_{x\to0^+}H(x)=1$, so the limit does not exist.
Yes; the right-hand limit equals $1$, so $H$ is continuous.
No; $H(0)$ is undefined.
Yes; $\lim_{x\to0}H(x)=1$ and $H(0)=1$.
Explanation
For continuity at x=0, H(0)=1, but left limit -1, right 1, DNE, not continuous. Sign function jump. Common: thinking point fixes limit issue. Jump discontinuity. Checklist: (1) Defined? (2) Limit? (3) Match?
Let $L(x)=\begin{cases}2,&x\ne-2\\2,&x=-2\end{cases}$. Is $L$ continuous at $x=-2$, and why?
No; $\lim_{x\to-2}L(x)=0$ but $L(-2)=2$.
No; $L(-2)$ is undefined.
Yes; $L(-2)=2$ so the limit is unnecessary.
Yes; $\lim_{x\to-2}L(x)=2$ and $L(-2)=2$.
No; $\lim_{x\to-2}L(x)$ does not exist because the function is constant.
Explanation
Continuity at x=-2: L(-2)=2, limit 2, continuous. Constant function. Omission: unnecessary worry over notation. Continuous. Use: (1) f(a)? (2) Limit? (3) Match?
For $c(x)=\begin{cases}1,&x<1\\2,&x\ge1\end{cases}$, is $c$ continuous at $x=1$, and why?
No; $\lim_{x\to1^-}c(x)=1$ and $\lim_{x\to1^+}c(x)=2$, so $\lim_{x\to1}c(x)$ does not exist.
Yes; the left-hand limit equals $1$, so it is continuous.
No; $c(1)$ is undefined.
Yes; $c(1)=2$ and $\lim_{x\to1}c(x)=2$.
No; $\lim_{x\to1}c(x)=1$ but $c(1)=2$.
Explanation
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).
Let $Z(x)=\begin{cases}x^2-4x+4,&x\ne2\\1,&x=2\end{cases}$. Is $Z$ continuous at $x=2$, and why?
Yes; $\lim_{x\to2}Z(x)=0$ and $Z(2)=0$.
No; $\lim_{x\to2}Z(x)$ does not exist because it is quadratic.
Yes; $Z(2)=1$ so the limit must be $1$.
No; $Z(2)$ is undefined.
No; $\lim_{x\to2}Z(x)=0$ but $Z(2)=1$.
Explanation
A function is continuous at x = a if all three conditions hold: f(a) defined, limit exists, and they equal. For Z(x) at x = 2, Z(2) = 1, but the limit of (x-2)² is 0, which does not equal 1, so discontinuous. Students commonly omit comparing the limit to the redefined value, thinking the limit alone suffices. Here, the quadratic approaches 0, but the point is set to 1, creating a discontinuity. Redefining doesn't ensure continuity unless it matches the limit. 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).
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?
Yes; the left-hand limit equals $-1$, so it is continuous.
Yes; $\lim_{x\to0}o(x)=-1$ and $o(0)=-1$.
No; $\lim_{x\to0^-}o(x)=-1$ and $\lim_{x\to0^+}o(x)=1$, so the limit does not exist.
No; $\lim_{x\to0}o(x)=0$ but $o(0)=-1$.
No; $o(0)$ is undefined.
Explanation
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).
Let $p(x)=\begin{cases}\sqrt{x},&x\ge0\\0,&x<0\end{cases}$. Is $p$ continuous at $x=0$, and why?
No; $\lim_{x\to0}p(x)$ does not exist because $\sqrt{x}$ is undefined for $x<0$.
Yes; only the right-hand limit matters at $0$.
No; $\lim_{x\to0}p(x)=1$ but $p(0)=0$.
No; $p(0)$ is undefined.
Yes; $p(0)=0$ and both one-sided limits equal $0$.
Explanation
For continuity, check definition, limit existence, and equality. At x = 0 for p(x), p(0) = 0 (from √x), left limit 0 (constant), right √x → 0, so limit = 0 matches, continuous. People often forget that for domains like square root, left limit is from the extension. Here, the extension to 0 for x<0 makes left continuous. This shows how to extend functions 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).
Let $f(x)=\begin{cases}\sin x/x,&x\ne0\\1,&x=0\end{cases}$. Is $f$ continuous at $x=0$, and why?
No; $\lim_{x\to0}\sin x/x$ does not exist because $\sin 0=0$.
Yes; the limit exists, so continuity holds even if $f(0)$ were different.
Yes; $f(0)=1$ and $\lim_{x\to0}f(x)=1$, so $\lim_{x\to0}f(x)=f(0)$.
No; $f(0)$ is undefined since $\sin 0/0$ is undefined.
No; $\lim_{x\to0}f(x)=0$ but $f(0)=1$.
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).