- Topic
- Continuity
- Difficulty
- Standard
- Suggested time
- 4 min
- Marks
- 2
Continuity practice question 17
The graph of a function is shown. Of , , and , which set lists exactly the ones at which is not continuous?
Test each x-value in turn: f(a) must be defined, the limit as x → a must exist, and the two must be equal.
At x = 1 the graph changes direction sharply. Does it break there, or only bend?
Theory UBC Math 100 §2.3CLP-1 §1.6Key results
Worked solution
Strategy
Test each x-value against the definition of continuity: the value f(a) must exist, the limit as x → a must exist, and they must agree. One failure is enough.
Working
Both branches climb without bound beside the dashed line: an infinite discontinuity.
The one-sided limits differ, so the limit does not exist: a jump. f(−1) = 1 makes f continuous from the left only.
The two lines of the V meet at the corner (1, 1). The graph bends sharply but does not break, so f is continuous at 1.
The limit and the value both exist, but they differ: a removable discontinuity.
Answer
The answer is f is not continuous at x = −3, −1 and 3; it is continuous at x = 1.
Check
A pencil tracing the graph must lift at x = −3, −1 and 3, where the dot at (3, 5) sits off the line, but passes the corner at x = 1 without lifting.
Where this goes wrong
Counting the corner at x = 1, or missing x = 3. A corner is where f′ fails to exist, but the graph is unbroken there; at x = 3 the limit exists, but the filled point puts the value somewhere else.
Leans on
One-sided limits, from Limits
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