_MATH1013_1ABCD_2025 Subsection 4.3 (closed on 1 Nov)

Compute [math: limx→0(ln⁡(x+1))2xsin⁡4x]\displaystyle \lim _{x\to 0}{\frac{(\ln{(x+1)})^2}{x\sin{4x}}}. (Use I to stand for [math: ∞]\infty if needed.)