Check continuity at a domain endpoint
A curve that stops at an end
The curve is \(f(x) = \sqrt{9 - x^2}\), the upper half of a circle of radius 3, so the function has values only for \(-3 \le x \le 3\).
The orange dot at \((-3, 0)\) is the left end of the curve. Nothing is drawn to its left, because the function has no values there.
Press Approach −3 from the right. The moving orange dot on the curve approaches the end, and its height approaches the value \(f(-3) = \sqrt{0} = 0\).
Nothing can approach −3 from the left, so at an end the one side that has values is the only side to test. The right end at 3 is the same, approached from the left.
One side at each end
At the left end \(a\), the curve has values only to its right. At the right end \(b\), it has values only to the left.
Each faint dashed line marks the height that the curve approaches from inside. Move an end dot with its slider. When the dot is off its dashed line, a red bar joins it to the end of the curve.
The end passes the test when its dot is at the height that the curve approaches from inside.
Notation
At an end of the domain, the function has values on one side only. A two-sided limit needs both sides, so use the one side that exists.
At the left end \(a\), use the right-hand limit. It must equal the value.
\[\lim_{x \to a^+} f(x) = \cy{f(a)}\]At the right end \(b\), use the left-hand limit. It must equal the value.
\[\lim_{x \to b^-} f(x) = \cy{f(b)}\]If the equation holds, \(f\) is continuous from inside at that end: from the right at \(a\), or from the left at \(b\). Write a square bracket there.
If \(f\) has no value at the end, or the equation fails, the end is not part of an interval of continuity. Write a round bracket there.
When it fails
Press Lift the end dot off the curve. The curve approaches the same heights from inside, but each end dot moves off its height, so a red bar joins it to the curve.
The equation fails, so the function is not continuous from inside at these ends, and each end takes a round bracket.
Example 1: A root at its left end
State the interval on which \(g(x) = \sqrt{x + 6}\) is continuous.
The root needs \(x + 6 \ge 0\), so the domain is \(x \ge -6\). The teal bar under the axis marks this domain.
The value at the left end exists, and the orange dot shows it at height 0.
\[g(-6) = \sqrt{0} = 0\]Use the right-hand limit, because only the right side has values.
\[\lim_{x \to -6^+} \sqrt{x + 6} = 0\]The limit equals the value, so \(g\) is continuous from the right at \(-6\). The end takes a square bracket.
\[[-6, \infty)\]
Example 2: A closed interval with a dot off the curve
A function is defined only on \([0, 4]\). Let \(h(x) = 2x + 1\) for \(0 \le x < 4\), and \(h(4) = 3\). State where \(h\) is continuous.
At the left end, the right-hand limit equals the value \(h(0) = 1\), so \(h\) is continuous from the right at 0.
\[\lim_{x \to 0^+} (2x + 1) = 1 = h(0)\]At the right end, use the left-hand limit. The line approaches the height 9, which the open circle marks.
\[\lim_{x \to 4^-} (2x + 1) = 9\]A tempting wrong move: The left-hand limit at 4 exists, so \(h\) is continuous on \([0, 4]\).
\[[0, 4]\]A limit that exists is not enough. The orange dot at \(x = 4\) is at height 3, not 9, so a red bar of length 6 separates it from the open circle.
The limit 9 does not equal the value \(h(4) = 3\), so \(h\) is not continuous at 4 and the right end takes a round bracket.
\[[0, 4)\]
Practice
State the interval on which \(q(x) = \sqrt{5 - x}\) is continuous.
Find the domain first. Its end is on the right, so move the slider toward it from the left.
Solution
The root needs \(5 - x \ge 0\), so the domain is \(x \le 5\), which the teal bar marks.
\[x \le 5\]The value at the end is \(q(5) = \sqrt{0} = 0\). Only the left side has values, so use the left-hand limit.
\[\lim_{x \to 5^-} \sqrt{5 - x} = 0 = q(5)\]The limit equals the value, so the end takes a square bracket.
\[(-\infty, 5]\]
Questions
Use the graph of \(f\) for the first two questions. The function is defined only on \([-2, 4]\). A filled dot is a value. An open circle is a height that the curve approaches without a point there.
At \(x = -2\), which limit tests continuity?
- The right-hand limit
- The left-hand limit
- The two-sided limit
- No limit applies, because \(f(-2)\) does not exist.
Answer
The right-hand limit
Is \(f\) continuous from inside at \(x = 4\)?
- No. The left-hand limit is 5, but \(f(4) = 2\).
- Yes. The function has a value at 4.
- Yes. The left-hand limit exists.
- No. There is no left-hand limit at 4.
Answer
No. The left-hand limit is 5, but \(f(4) = 2\).
State the interval on which \(w(x) = \sqrt{3x - 12}\) is continuous.
- \([4, \infty)\)
- \((4, \infty)\)
- \([0, \infty)\)
- \((-\infty, \infty)\)
Answer
\([4, \infty)\)
Summary
At an end of the domain, you compare the value with the limit from inside: right-hand at a left end, left-hand at a right end.
This works because a function with no values on one side cannot approach anything from there, so the one-sided test is the whole test.
The test decides whether an end takes a square or a round bracket in an interval of continuity.
This is one of four skills in topic 1.12 of Which & Why: AP Calculus.