## Thomas' Calculus 13th Edition

$k$
We know $$\lim\limits_{\theta \to 0} \dfrac{\mathrm{sin}\left(\theta \right)}{\theta }=1.$$ Thus for any nonzero real number $a$, it follows that $$\underset{\theta \to 0}{\lim}\dfrac{\mathrm{sin}\left(a\theta \right)}{a\theta }=1.$$ We will use this fact to compute the limit below. So to see that this is true, we let $x=a\theta.$ Then since $x \to 0$ as $\theta \to 0$, we have $$\underset{\theta \to 0}{\lim}\dfrac{\mathrm{sin}\left(a\theta \right)}{a\theta }=\underset{x\to 0}{\lim}\dfrac{\mathrm{sin}\left(x\right)}{x}=1.$$ Now, we want to find $\underset{t\to 0}{\lim}\dfrac{\mathrm{sin}\left(kt\right)}{t}$, where $k$ is a constant. We first suppose $k=0$. Then $$\underset{t\to 0}{\lim}\dfrac{\mathrm{sin}\left(kt\right)}{t}=\underset{t\to 0}{\lim}\dfrac{\mathrm{sin}\left(0\right)}{t}=\underset{t\to 0}{\lim}0=0=k.$$ Now suppose $k\neq 0$. Then $$\underset{t\to 0}{\lim}\dfrac{\mathrm{sin}\left(kt\right)}{t}=\underset{t\to 0}{\lim}\dfrac{k\mathrm{sin}\left(kt\right)}{kt}=k\left(\underset{t\to 0}{\lim}\dfrac{\mathrm{sin}\left(kt\right)}{kt}\right)=k\left(1\right)=k.$$ Thus for any constant $k$, $\underset{t\to 0}{\lim}\dfrac{\mathrm{sin}\left(kt\right)}{t}=k$.