Past exam of the mathematics course of the University of Oxford / 2017 / Prelims / Mathematics II / 6 / c

Table of contents
- Solution c

Solution

No. On

(- 1, 1)

take

c = 0

and

f (x) = {x^{2} sin (1/ x), 0, x \neq = 0, x = 0.

At zero, its difference quotient has absolute value

∣ x sin (1/ x) ∣ \leq ∣ x ∣ \to 0

, so

f^{'} (0) = 0

. Away from zero, the product rule and chain rule apply. Define

A_{n} = 2 πn, B_{n} = {π /2, 3 π /2, n even, n odd, x_{n} = \frac{1}{A _{n}}, y_{n} = \frac{1}{A _{n} + B _{n}} (n \geq 1) .

These are distinct positive inputs tending to zero. Since

sin A_{n} = 0

and

sin (A_{n} + B_{n}) = sin B_{n}

,

\frac{f ( y _{n} ) - f ( x _{n} )}{y _{n} - x _{n}} = - \frac{s i n B _{n}}{B _{n}} \frac{A _{n}}{A _{n} + B _{n}} .

The even-indexed subsequence tends to

- 2/ π

, whereas the odd-indexed subsequence tends to

2/ (3 π)

. Thus the limit of a sequence does not exist. The two points lie on the same side of

c

, so the nonnegative-weight estimate in part (b) is unavailable. This is a concrete instance of differentiability does not imply continuity of the derivative.

 Ancestors (11)

 View article source

 Discussion (0)

There are no discussions about this article yet.

 Articles by others on the same topic (0)

There are currently no matching articles.

  See all articles in the same topic Create my own version