Check the three continuity conditions
A curve that reaches its point
The curve is the graph of \(f(x) = x^2 - 2x\), and the filled dot is its point \((3, 3)\).
Press Close in from the left. An orange point moves along the curve toward \(x = 3\). At \(x = 2.9\) its height is 2.61, and at \(x = 2.99\) its height is about 2.96.
Press Close in from the right. At \(x = 3.1\) the height is 3.41, and at \(x = 3.01\) the height is about 3.04.
From both sides, the height approaches 3, which is the height of the filled dot. The curve arrives at the dot with no gap.
The same story at any input
The filled dot is the point \((c, f(c))\), and the dashed orange line marks its height, the value \(\cy{f(c)}\).
Drag the first slider to move \(c\). Drag the second slider to close in on \(c\). The heights of the two orange points approach the dashed line.
The function is continuous at \(c\) when both sides approach the height of the dot, so the curve arrives at the dot with no gap.
Notation
A function is continuous at \(x = c\) when it passes three tests. Test 3 compares the value with the limit, so you check first that both exist.
Test 1: the value \(\cy{f(c)}\) exists. The filled dot is on the graph above \(x = c\).
\[\cy{f(c)} \text{ exists}\]Test 2: the limit exists. The orange points show that the curve approaches the same height \(\cy{L}\) from the left and from the right.
\[\lim_{x \to c} f(x) = \cy{L}\]If the two sides approach different heights, there is no limit.
Test 3: the limit equals the value. The curve approaches the height of the dot, so it arrives at the dot.
\[\lim_{x \to c} f(x) = \cy{f(c)}\]If one test fails, the function is not continuous at c. A passing test does not repair a failing one.
When it fails
Press Remove the dot. The curve still approaches a height, but no value exists, so test 1 fails. Press Lift the dot. The red bar is the gap between the limit and the value, so test 3 fails.
Press Split the sides. The two sides approach different heights, so test 2 fails. The dot cannot repair this, because it sits at one height only.
Example 1: a function that passes all three tests
Let \(f(x) = \dfrac{x^2 - 16}{x - 4}\) for \(x \ne 4\), and let \(f(4) = 8\). Is \(f\) continuous at \(x = 4\)?
Test 1: the problem gives \(f(4) = 8\), so the value exists. The filled dot is at height 8.
\[f(4) = 8\]For \(x \ne 4\), factor the top. The factor \(x - 4\) cancels, because \(x\) is never equal to 4 while it approaches 4.
\[\begin{aligned} \frac{x^2 - 16}{x - 4} &= \frac{(x - 4)(x + 4)}{x - 4} \\ &= x + 4 \end{aligned}\]Test 2: near 4, the function equals \(x + 4\), which approaches 8 from both sides, so the limit exists.
\[\lim_{x \to 4} f(x) = \cy{8}\]Test 3: the limit 8 equals the value \(f(4)\). The dot lies on the curve, so all three tests pass and \(f\) is continuous at 4.
\[\lim_{x \to 4} f(x) = 8 = f(4)\]
Example 2: a function that fails
Let \(g(x) = \dfrac{x^2 - 5x + 6}{x - 2}\) for \(x \ne 2\), and let \(g(2) = 3\). Is \(g\) continuous at \(x = 2\)?
Test 1: the problem gives \(g(2) = 3\), so the value exists. The filled dot is at height 3.
\[g(2) = 3\]For \(x \ne 2\), the factor \(x - 2\) cancels, so \(g(x) = x - 3\).
\[\begin{aligned} \frac{x^2 - 5x + 6}{x - 2} &= \frac{(x - 2)(x - 3)}{x - 2} \\ &= x - 3 \end{aligned}\]Test 2: near 2, the function equals \(x - 3\), which approaches \(-1\) from both sides, so the limit exists.
\[\lim_{x \to 2} g(x) = \cy{-1}\]A tempting wrong move: The limit exists, so \(g\) is continuous at 2.
\[\lim_{x \to 2} g(x) = -1 \Rightarrow \text{continuous}\]A limit that exists passes only test 2. The curve approaches height −1, but the filled dot is at height 3, and the red bar of length 4 is the gap between them.
Test 3 fails, because the limit \(-1\) differs from the value 3, so \(g\) is not continuous at 2.
\[-1 \ne 3\]
Practice
Let \(h(x) = \dfrac{|x + 2|}{x + 2}\) for \(x \ne -2\), and let \(h(-2) = 1\). Which test fails at \(x = -2\)?
Find the height of the graph close to \(-2\) on the right and on the left. Compare the two heights in the picture.
Solution
Test 1: the problem gives \(h(-2) = 1\), so test 1 passes. The filled dot is at height 1.
If \(x > -2\), then \(|x + 2| = x + 2\), so \(h(x) = 1\). The right side approaches 1.
\[\lim_{x \to -2^+} h(x) = 1\]If \(x < -2\), then \(|x + 2| = -(x + 2)\), so \(h(x) = -1\). The left side approaches \(-1\).
\[\lim_{x \to -2^-} h(x) = -1\]The sides approach different heights, so the limit does not exist and test 2 fails, even though the value exists.
Questions
Use the graph of F for the first two questions. A filled dot is a value. An open ring is a height that the curve approaches with no point there.
At \(x = 3\), which test fails?
- Test 1: the value exists
- Test 2: the limit exists
- Test 3: the limit equals the value
- None. The function is continuous at 3.
Answer
Test 3: the limit equals the value
At \(x = 5\), which statement is correct?
- The value exists and the limit exists, but they differ.
- The value exists, but the limit does not exist.
- The limit is 3, because the dot is at height 3.
- The value does not exist.
Answer
The value exists, but the limit does not exist.
Let \(q(x) = \dfrac{x^2 - 9}{x + 3}\) for \(x \ne -3\), and let \(q(-3) = -6\). Is \(q\) continuous at \(x = -3\)?
- Yes. The limit is \(-6\), which equals \(q(-3)\).
- No. The denominator is 0 at \(x = -3\).
- No. The limit is 6, which differs from \(q(-3)\).
Answer
Yes. The limit is \(-6\), which equals \(q(-3)\).
Summary
You test continuity at a point in order: the value exists, the limit exists, and the two are equal.
This works because a continuous curve arrives at its dot with no gap, and each test catches one way a gap can appear.
Use the tests at one input when a question asks whether a function is continuous there.
This is one of four skills in topic 1.11 of Which & Why: AP Calculus.