$\lim_{x\to 0}\frac{\sin 3x}{x}$
$=\lim_{x\to 0}\frac{3\sin 3x}{3x}=3\cdot 1=3$
$\lim_{x\to 0}\frac{\sin 5x}{\sin 2x}$
$=\frac{\sin 5x}{5x}\cdot\frac{2x}{\sin 2x}\cdot\frac{5}{2}=1\cdot 1\cdot\frac{5}{2}=\frac{5}{2}$
$\lim_{x\to 0}\frac{1-\cos x}{x^2}$
استخدم $1-\cos x=2\sin^2\frac{x}{2}$:
$\frac{2\sin^2(x/2)}{x^2}=\frac{1}{2}\left(\frac{\sin(x/2)}{x/2}\right)^2 \to \frac{1}{2}$