- Topic
- Continuity
- Difficulty
- Standard
- Suggested time
- 4 min
- Marks
- 2
Continuity practice question 18
The graph of a function is shown. is continuous at when three conditions hold: first, is defined; second, exists; third, the two are equal. At each of and , which condition is the first to fail?
At each point, look first for a filled point at that x-value: is the value defined? Then follow the graph towards the point from each side: do the two sides approach the same height?
At , check the conditions in order, starting with whether a filled point sits anywhere above or below the open circle. At , is the filled point; read off the height each piece approaches from its own side.
Theory UBC Math 100 §2.3CLP-1 §1.6Key results
Worked solution
Strategy
Take the conditions in order at each point. The value comes from a filled point at that x-value, and the limit from the heights the graph approaches on each side.
Working
There is no filled point at x = −2, only the open circle, which marks a missing point. The first condition fails.
The limit does exist: from both sides the curve approaches the open circle at height 1. Defining g(−2) = 1 would remove the discontinuity.
The filled point gives the value, so the first condition holds.
The one-sided limits disagree, so the limit does not exist and the second condition fails: a jump.
Answer
The answer is at x = −2, g(−2) is not defined; at x = 1, the limit does not exist.
Check
Defining g(−2) = 1 would fill the hole and make g continuous there, but no value of g(1) can join heights 4 and 2.
Where this goes wrong
Treating the hole at x = −2 as a missing limit. The limit there exists and equals 1; what is missing is the value. At x = 1 it is the other way round: the value exists, but the two sides approach different heights.
Leans on
One-sided limits, from Limits
Try a similar question
A function is continuous at , and . Which of the following must be true?Saved in this browser only.