Finding the Limit of a Trigonometric Function

Find:

1. \(\lim\limits_{\theta \rightarrow 0}\dfrac{\sin (3\theta)}{\theta}\)
2. \(\lim\limits_{\theta \rightarrow 0}\dfrac{\sin (5\theta)}{\sin (2\theta)}\)

Solution (a) Since \(\lim\limits_{\theta \rightarrow 0}\dfrac{\sin (3\theta)}{\theta}\) is not in the same form as \(\lim\limits_{\theta\rightarrow 0}\dfrac{\sin \theta}{\theta}\), we multiply the numerator and the denominator by 3, and make the substitution \(t = 3\theta\). \[ \begin{equation*} \lim_{\theta \rightarrow 0}\frac{\sin (3\theta )}{\theta }=\lim_{\theta \rightarrow 0}\frac{3\sin (3\theta )}{3\theta } \underset{\color{#0066A7}{{t\rightarrow 0}{\hbox{ as }}{\theta \rightarrow 0}}}{\underset{\color{#0066A7}{{t=3\theta}}}{\underset{\color{#0066A7}{\uparrow }}{=}}} \lim_{t\rightarrow 0}\left( 3\frac{\sin t}{t}\right) =3\lim_{t\rightarrow 0} \frac{\sin t}{t}\underset{\color{#0066A7}{{\lim\limits_{t\rightarrow 0}\dfrac{\sin t}{t}=1 }}}{\underset{\color{#0066A7}{\uparrow }}{=}}\!\!(3)(1)=3 \end{equation*} \]

(b) We begin by dividing the numerator and the denominator by \(\theta\). Then \[ \begin{equation*} \dfrac{\sin (5\theta)}{\sin (2\theta)} = \dfrac{ \dfrac{\sin (5\theta)}{\theta}}{\dfrac{\sin (2\theta)}{\theta}} \end{equation*} \]

Now we follow the approach in (a) on the numerator and on the denominator. \[ \begin{eqnarray*} \lim\limits_{\theta \rightarrow 0}\dfrac{\sin ( 5\theta ) }{ \theta } &=&\lim\limits_{\theta \rightarrow 0}\dfrac{5\sin ( 5\theta ) }{5\,\theta } \underset{\color{#0066A7}{{t\rightarrow 0\hbox{ as }\theta \rightarrow 0}}} {\underset{\color{#0066A7}{t=5\theta}}{\underset{\color{#0066A7}{\uparrow }}{=}}}\lim\limits_{t\rightarrow 0} \dfrac{5\sin t}{t}=5\lim\limits_{t\rightarrow 0}\left( \dfrac{\sin t}{t} \right) =5 \\ \lim\limits_{\theta \rightarrow 0}\dfrac{\sin ( 2\theta ) }{ \theta } &=&\lim\limits_{\theta \rightarrow 0}\dfrac{2\sin ( 2\theta ) }{2\theta } \underset{\color{#0066A7}{{t\rightarrow 0\hbox{ as }\theta \rightarrow 0}}}{ \underset{\color{#0066A7}{{t=2\theta}}}{\underset{\uparrow }{=}}}\lim\limits_{t\rightarrow 0}\left( \dfrac{2\sin t}{t}\right) =2\lim\limits_{t\rightarrow 0}\dfrac{\sin t}{t}=2 \\ \lim\limits_{\theta \rightarrow 0} \dfrac{\sin (5\theta)}{\sin (2\theta)} &=&\dfrac{ \lim\limits_{\theta \rightarrow 0}\dfrac{\sin (5\theta)}{\theta}}{\lim\limits_{\theta \rightarrow 0}\dfrac{\sin (2\theta)}{\theta}} = \frac {5}{2} \end{eqnarray*} \]