To check continuity at 0 we need
f(x)=f(0)=0.
We'll use
|sin‌(⋅)|≤1 so
|f(x)|≤|xα| and see when that goes to 0 .
To check differentiability at 0 , we use
f′(0)=‌=xα−1sin‌(‌)and again compare
|xα−1| to 0 .
Short Solution
1. Continuity at
x=0For
x≠0,
f(x)=xαsin‌(‌)Use
|sin‌(θ)|≤1 :
|f(x)|≤|xα|As
x⟶0,
- If
α>0, then
|xα|⟶0⇒f(x)⟶0, so
f is continuous at 0 .
- If
α≤0, then
|xα| does not go to 0 (it either stays 1 or blows up), so the limit is not 0 .
Thus, continuity at 0 holds iff
α>0, and this doesn't depend on
β.
So:
- Statement (2) is false (cannot hold for all
α∈R ).
- Statement (3) is too restrictive (it unnecessarily forces
β>0 ).
- Statement (4) matches: continuous at 0 for all
c>0 and any real
β.
2. Differentiability at
x=0For completeness, check (1):
f′(0)=‌=xα−1sin‌(‌).Using
|sin‌|≤1 :
|xα−1sin‌(‌)|≤|xα−1|.This goes to 0 only if
α−1>0⇒α>1, so differentiability at 0 is not true for all
α>0.
So (1) is false.