 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/19/a/i/solution

Since

A > 0

and

ω > 0

are fixed, the standard sine limit gives

f (x) = A ω sgn (x) ∣ x ∣^{p + 1} \frac{s i n ( ω x )}{ω x} \sim A ω sgn (x) ∣ x ∣^{p + 1} (x \to 0) .

p > - 1

, this tends to the assigned value zero. At

p = - 1

, the right and left limits are

+ A ω

and

- A ω

; if

p < - 1

, they diverge with opposite signs. Hence

f is continuous at 0 ⟺ p > - 1, for every allowed A, ω > 0.

The cancellation comes from the simple zero of

sin (ω x)

at the origin, not rapid oscillation there.

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