Defining Continuity at a Point
Introduced| English |
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| continuous/kənˈtɪnjuːəs/ |
"Draw it without lifting your pen"
- Intuitively, a function is continuous 连续 at $c$ if its graph has no break there — no hole, no jump, no blow-up.
- But "no break" needs a precise test you can check, not just eyeball.
- Calculus pins it to three conditions, all about the point $c$.
- Pass all three and the curve flows smoothly through $\big(c, f(c)\big)$.
A curve with no break
y = ax² + bx + c
This polynomial passes all three continuity conditions at every point — defined, limit exists, and the two match.
The three conditions
- (1) $f(c)$ is defined — the point actually exists (no hole).
- (2) $\displaystyle\lim_{x\to c}f(x)$ exists — both one-sided limits agree (no jump, no blow-up).
- (3) They are equal: $\displaystyle\lim_{x\to c}f(x)=f(c)$ — the curve arrives exactly where the point sits.
- All three must hold. Continuity at $c$ is precisely conditions (1) + (2) + (3).
Select all conditions required for $f$ to be continuous at $c$.
The three conditions are defined, limit-exists, and equal. Being a polynomial is not required.
The third continuity condition says the limit must ____ the function value $f(c)$.
$\lim_{x\to c}f(x)=f(c)$.
Break any one → discontinuity
- Fail (1): the point is missing — a hole even if the limit exists.
- Fail (2): the sides disagree or the function blows up — a jump or infinite break.
- Fail (3): both exist but the dot floats off the curve — a removable hole plotted elsewhere.
- Each failure matches a discontinuity type from the previous lesson.
If $f(c)=1$ and $\lim_{x\to c}f(x)=4$, then at $c$ the function is...
Both exist but disagree — condition 3 fails, so $f$ is discontinuous (a removable hole).
Checking a specific point
- Plug in: is $f(c)$ a real number? (Condition 1.)
- Take the limit from both sides: do they agree on a value $L$? (Condition 2.)
- Compare: does $L=f(c)$? (Condition 3.)
- Only "yes, yes, yes" earns the label continuous at $c$.
Merely knowing $f(c)$ is defined is enough to conclude $f$ is continuous at $c$.
You also need the limit to exist and to equal $f(c)$.
For $f(x)=x^2+1$, is $f$ continuous at $x=2$?
All three conditions hold, so $f$ is continuous at $2$ (polynomials are continuous everywhere).
For continuity of $f$ at $x=1$ we need $\lim_{x\to1}f(x)=f(1)$. If $\lim_{x\to1}f(x)=7$, what must $f(1)$ equal?
Condition 3 forces $f(1)=7$.
All three conditions are needed — checking only that $f(c)$ exists, or only that the limit exists, is not enough. A function can have a perfectly good $f(c)=1$ and a perfectly good $\lim_{x\to c}f(x)=4$, yet still be discontinuous because $1\neq4$ (condition 3 fails).
Is $f(x)=\begin{cases}\dfrac{x^2-1}{x-1},&x\neq1\\[4pt]3,&x=1\end{cases}$ continuous at $x=1$?
- (1) $f(1)=3$ — defined. ✓
- (2) $\displaystyle\lim_{x\to1}\dfrac{x^2-1}{x-1}=\lim_{x\to1}(x+1)=2$ — exists. ✓
- (3) $2 \neq 3$ — not equal. ✗
- Condition (3) fails, so $f$ is discontinuous at $1$ (a removable hole with the point plotted at height $3$).
$f$ is continuous at $c$ exactly when three things hold: $f(c)$ is defined, $\displaystyle\lim_{x\to c}f(x)$ exists, and the two are equal. Fail any one and you get a discontinuity — the failed condition tells you which type.