Final answer:
To prove that lim (x)¹/³ * sin (1/x²) = 0 using the squeeze theorem, we can show that the function is bounded between two other functions that converge to 0 as x approaches 0.
Step-by-step explanation:
To prove that lim (x)¹/³ * sin (1/x²) = 0 using the squeeze theorem, we need to show that the function is bounded between two other functions that converge to 0 as x approaches 0.
First, we can notice that -1 ≤ sin(1/x²) ≤ 1 for all x ≠ 0.
This means that -x(1/x²) ≤ x(1/x²) * sin(1/x²) ≤ x(1/x²).
As x approaches 0, both of these inequalities converge to 0.
Therefore, by the squeeze theorem, we can conclude that lim (x)¹/³ * sin (1/x²) = 0.