Explain why continuity fails
Four breaks, four reasons
A break is an input where the function is not continuous, and this graph has four. A filled dot marks a value, and an open ring marks a height that the curve approaches. Press a button to move the orange line to a break.
Test 1 asks whether the value exists. Test 2 asks whether the limit exists. Test 3 asks whether the limit equals the value.
At \(x = -4\) there is no filled dot, so test 1 fails. At \(x = -1\) the sides approach the heights 1 and 3, so there is no limit and test 2 fails.
At \(x = 2\) the curve grows without bound on both sides, so test 2 fails, although the dot at height 0 exists. At \(x = 5\) both sides approach 4, but the dot is at 1, so test 3 fails.
Every break fails a test
Each button changes this graph at the input c. The orange line marks the height that the curve approaches. If you remove the dot, test 1 fails. If you lift the dot off the curve, test 3 fails.
If you split the sides or make the curve unbounded, test 2 fails. One failed test is enough for a break, and the tests that pass do not repair it.
Notation
To explain why \(f\) is not continuous at \(c\), name the test that fails.
Test 1 fails when \(\cy{f(c)}\) does not exist, so the graph has no filled dot above \(x = c\).
\[\cy{f(c)} \text{ does not exist}\]Test 2 fails when the limit does not exist. The sides approach different heights, the curve grows without bound, or it oscillates.
\[\lim_{x \to c} f(x) \text{ does not exist}\]Test 3 fails when the value and the limit both exist but differ. The red bar shows the gap between the value and the limit.
\[\lim_{x \to c} f(x) \ne \cy{f(c)}\]The tests use only values and limits, so the slopes on the two sides of \(c\) play no part.
When it fails
Press Split the sides. The graph has a jump and no dot, so tests 1 and 2 fail. Press Put a dot at the break. The new dot is at one of the two heights, so test 1 passes.
Test 2 still fails, because the two sides approach different heights. A value does not repair a missing limit.
Example 1: a curve that grows without bound
Let \(f(x) = \dfrac{1}{(x - 3)^2}\) for \(x \ne 3\), and let \(f(3) = 0\). Explain why \(f\) is not continuous at \(x = 3\).
The filled dot shows the value \(f(3) = 0\), so test 1 passes.
\[f(3) = 0\]Near 3, \((x - 3)^2\) is a small positive number, so \(f(x)\) is a large positive number. At \(x = 3.01\) the height is 10,000, far above the figure.
\[f(3.01) = \frac{1}{0.01^2} = 10000\]From both sides, the heights grow without bound and approach no single height, so the limit does not exist and test 2 fails.
\[\lim_{x \to 3} f(x) = \infty\]The symbol ∞ describes the unbounded growth and is not a number.
The function is not continuous at 3. The dot at height 0 does not repair the missing limit.
Example 2: a corner is not a break
Let \(g(x) = x^2\) for \(x \le 2\), and let \(g(x) = 3x - 2\) for \(x > 2\). The pieces have different slopes at \(x = 2\). Does \(g\) fail a test there?
The rule for \(x \le 2\) gives \(g(2) = 4\), so test 1 passes.
\[g(2) = 2^2 = 4\]The left piece approaches \(2^2 = 4\) and the right piece approaches \(3(2) - 2 = 4\), so the limit is 4 and test 2 passes.
\[\lim_{x \to 2} g(x) = 4\]A tempting wrong move: The slope of \(x^2\) at 2 is 4, and the slope of \(3x - 2\) is 3. The slopes differ, so \(g\) is not continuous at 2.
\[4 \ne 3\]The dashed orange line shows that the slopes differ, but the graph is one connected curve through (2, 4) with no gap. The tests ask about heights, and the heights agree.
The limit 4 equals the value \(g(2) = 4\), so test 3 passes. All three tests pass, so \(g\) is continuous at 2.
Practice
Let \(p(x) = \sin\!\left(\dfrac{1}{x}\right)\) for \(x \ne 0\), and let \(p(0) = 0\). Which test fails at \(x = 0\), and why?
Use the slider to find inputs near 0 where the graph is at height 1. Then find inputs near 0 where it is at height \(-1\).
Solution
The filled dot shows the value \(p(0) = 0\), so test 1 passes.
At \(x = \tfrac{2}{9\pi}\) the height is \(\sin\left(\tfrac{9\pi}{2}\right) = 1\). At \(x = \tfrac{2}{11\pi}\) the height is \(\sin\left(\tfrac{11\pi}{2}\right) = -1\).
\[p\!\left(\tfrac{2}{9\pi}\right) = 1, \quad p\!\left(\tfrac{2}{11\pi}\right) = -1\]For each whole number \(n\), the input \(\tfrac{2}{(4n + 1)\pi}\) gives height 1 and the input \(\tfrac{2}{(4n + 3)\pi}\) gives height \(-1\). As \(n\) grows, both inputs approach 0.
Near 0 the curve oscillates between 1 and −1 and approaches no single height, so the limit does not exist and test 2 fails.
Questions
Use the graph of f for the first two questions.
At x = 0 the graph has a corner. Which statement is correct?
- f is not continuous, because the slopes on the two sides differ.
- f is continuous, because all three tests pass.
- Test 2 fails, because the two sides rise at different rates.
Answer
f is continuous, because all three tests pass.
At \(x = 3\), which test fails, and why?
- Test 1, because \(f(3)\) does not exist.
- Test 2, because the left side approaches 4 and the right side approaches 1.
- Test 3, because the limit is 1 and the value is 4.
Answer
Test 2, because the left side approaches 4 and the right side approaches 1.
Let \(u(x) = \dfrac{1}{x + 1}\) for \(x \ne -1\), and let \(u(-1) = 0\). Which test fails at \(x = -1\)?
- Test 1
- Test 2
- Test 3
Answer
Test 2
Summary
You explain a break by naming the test that fails: the value is missing, the limit does not exist, or the two differ.
This works because the tests use heights only, so a corner never breaks continuity and a value never repairs a missing limit.
Use it when a question asks why a function is not continuous at a given input.
This is one of four skills in topic 1.11 of Which & Why: AP Calculus.