- Difficulty
- Capstone
- Suggested time
- 10 min
- Marks
- 5
The definition of the derivative practice question 26
Let for and . Using the definition of the derivative, show that exists and find its value.
on its own is badly behaved near 0. What does the extra factor do to the difference quotient at 0?
The difference quotient at 0 is .
has no limit as , but it stays between and . So : what does tend to?
Theory UBC Math 100 §3.3CLP-1 §2.2Key results
Worked solution
Strategy
Nothing but the definition works here, because the formula has no derivative rule to apply at 0. The difference quotient reduces to h·sin(1/h), a bounded factor times h, and h tends to 0.
Working
Set up the definition at 0; one power of h cancels.
Because |sin(1/h)| ≤ 1 for every h ≠ 0.
A quantity whose size is at most |h| is forced to 0 along with |h|.
The limit that defines the derivative exists and equals 0.
Answer
The answer is the derivative at 0 exists and equals 0.
What earns the marks
- Writes the difference quotient at 0 and simplifies it to h·sin(1/h), noting that f(0) = 0.
- Bounds the quotient: |h·sin(1/h)| ≤ |h|, because |sin(1/h)| ≤ 1.
- Concludes that the quotient tends to 0 because |h| does, so f′(0) exists and equals 0.
Check
The graph is squeezed between y = x² and y = −x², both flat at 0, so its slope at 0 is trapped at 0 too.
Where this goes wrong
Declaring that f′(0) does not exist because sin(1/h) oscillates. The oscillation is real, but it is multiplied by h, which crushes it: its size is at most |h|, and |h| tends to 0.
Leans on
Bounded times vanishing, from Limits
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