## Calculus (3rd Edition)

We set $\theta=4h$. As $h\rightarrow0$, $\theta$ which is the multiple of $h$ also tends to $0$. Therefore, we can write $\lim\limits_{h \to 0}\frac{\sin 4h}{4h}=\lim\limits_{\theta \to 0}\frac{\sin \theta}{\theta}=1$ (from theorem 2)