- Topic
- Continuity
- Difficulty
- Standard
- Suggested time
- 3 min
- Marks
- 2
Continuity practice question 14
For which value of the constant is the function below continuous at ?
oscillates faster and faster as and has no limit, but it never leaves the interval . What does that say about the size of ?
for every . What happens to as ?
Theory UBC Math 100 §2.3CLP-1 §1.6Key results
Worked solution
Strategy
Continuity at 0 needs k to equal the limit of f at 0. The sine factor has no limit, but it is bounded, and it is multiplied by x, which tends to 0: a bounded factor times a factor tending to 0 tends to 0.
Working
The sine of any number lies between −1 and 1.
Multiply by |x|: the oscillation is trapped in a band that narrows to 0.
The distance from f(x) to 0 is at most |x|, and that tends to 0.
Continuity: the value must equal the limit.
Answer
The answer is k equals 0.
Check
At x = 0.01 and x = 0.001, f is about −0.0087 and 0.00093: small and shrinking with x, whatever the sine does.
Where this goes wrong
Giving up because sin(2/x) has no limit, or treating x sin(2/x) as the standard limit sin u/u and answering 1 or 2. The oscillation is bounded and multiplied by x, which tends to 0, so the product tends to 0.
Leans on
Bounded times vanishing, from Limits
Try a similar question
Evaluate the limit.Saved in this browser only.