Wiskunde Multiple Choice Vraag

$\lim_{x\to\,0}\;\frac{x^3-\sin^3x}{\sin^5x}=$	A. 1
B. \(\boldsymbol{\frac35}\)
C. \(\boldsymbol{\frac53}\)
D. \(\boldsymbol{\frac12}\)
E. \(\boldsymbol{\frac15}\)

$\inline\\\displaystyle\lim_{x\to\,0}\;\frac{x^3-\sin^3x}{\sin^5x}=\lim_{x\to\,0}\;\frac{(x-\sin x)(x^2+x\sin x+\sin^2x)}{\sin^3x\cdot\sin^2x}\\=\lim_{x\to\,0}\;\frac{x-\sin x}{\sin^3x}\cdot\lim_{x\to\,0}\;\frac{x^2+x\sin x+\sin^2x}{\sin^2x}\\=\lim_{x\to\,0}\;\frac{x-\sin x}{\sin^3x}\cdot\left(\lim_{x\to\,0}\;\frac{x^2}{\sin^2x}+\lim_{x\to\,0}\;\frac{x}{\sin x}+1\right)\\\left(=\frac00\right)\cdot\left(\lim_{x\to\,0}\;\frac{x^2}{\sin^2x}+\lim_{x\to\,0}\;\frac{x}{\sin x}+1\right)\\\overset{H}{=}\lim_{x\to\,0}\;\frac{1-\cos x}{3.\sin^2x.\cos x}\cdot(1^2+1+1)\\=3\cdot\lim_{x\to\,0}\;\frac{1}{3.(1+\cos x).\cos x}=3\cdot\frac{1}{3.(1+1).1}=\frac12$