$\displaystyle\lim_{h\longrightarrow 0} \dfrac{\sin h}{h}=1$ and $\displaystyle\lim_{h\longrightarrow 0} \dfrac{\cos h - 1}{h}=0$

If we take an angle $h$, the quantities $\dfrac{\sin h}{h}$ and $\dfrac{1-\cos h}{h}$ (note we have taken the opposite, $1-\cos h=-(\cos h - 1)$) are respectively the green curve divided by the blue curve and the red curve divided by the blue curve

When $h\longrightarrow 0$, green and blue curves tend to be equal in length, whereas the red one is much smaller than the blue one with a ratio that tends to zero. So we have $$\dfrac{\sin h}{h}\longrightarrow 1 \qquad \dfrac{\cos h -1}{h}=-\dfrac{1-\cos h}{h}\longrightarrow 0$$

When $h\longrightarrow 0$, green and blue curves tend to be equal in length, whereas the red one is much smaller than the blue one with a ratio that tends to zero. So we have $$\dfrac{\sin h}{h}\longrightarrow 1 \qquad \dfrac{\cos h -1}{h}=-\dfrac{1-\cos h}{h}\longrightarrow 0$$