\(\text{We steunen op}\;\displaystyle\lim_{x\to0}\frac{\sin ax}{bx}=\lim_{x\to0}\frac{ax}{\sin bx}=\frac ab\\ \displaystyle\lim_{x\to0}\:\frac{1-\cos3x}{1-\frac{\sin x}{x}}=\lim_{x\to0}\:\frac{x-x.\sin3x}{x-\sin x}\left(=\frac00\right)\\\overset{H}{=}\displaystyle\lim_{x\to0}\:\frac{1-\cos3x+3x\sin x}{1-\cos x}\left(=\frac00\right)\\\overset{H}{=}\displaystyle\lim_{x\to0}\:\frac{0+3\sin3x+3\sin3x+9x\cos3x}{\sin x}\\=3\displaystyle\lim_{x\to0}\:\left(\frac{\sin3x}{x}+\frac{\sin3x}{x}+\frac{\sin3x}{x}.\cos x\right)\\=3(3+3+3.1)=3.9=27\) ^{gricha - v7078 - 31.3.2026}