Consider the limit
L=x→0limx21−cos(4x).
The expression appears to be indeterminate of the form 00.
a. Verify algebraically that 1−cos(4x)=2sin2(2x).
b. Rewrite L entirely in terms of sin(2x) and x, showing each algebraic step clearly.
c. Explain, without using L’Hôpital’s Rule, why the form obtained in part (b) guarantees that L approaches a finite, non-zero value as x→0. State that value explicitly.