## University Calculus: Early Transcendentals (3rd Edition)

$$\lim_{t\to0}\frac{\sin kt}{t}=k$$
$$A=\lim_{t\to0}\frac{\sin kt}{t}$$ To be able to apply Theorem 7, we need $kt$ in the denominator, instead of $t$. So we would multiply $\frac{\sin kt}{t}$ with $\frac{k}{k}$: $$A=\lim_{t\to0}\Big(\frac{k}{k}\times\frac{\sin kt}{t}\Big)=k\lim_{t\to0}\frac{\sin kt}{kt}$$ Take $\theta=kt$. Then as $t\to0$, $\theta\to0\times k=0$ $$A=k\lim_{\theta\to0}\frac{\sin\theta}{\theta}$$ Apply Theorem 7: $$A=k\times1=k$$