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### Section 3-5 : Derivatives of Trig Functions

2. Evaluate $$\displaystyle \mathop {\lim }\limits_{\alpha \to \,0} \frac{{\sin \left( {12\alpha } \right)}}{{\sin \left( {5\alpha } \right)}}$$ .

Show Solution

All we need to do is set this up to allow us to use the fact from the notes in this section.

\begin{align*}\mathop {\lim }\limits_{\alpha \to \,0} \frac{{\sin \left( {12\alpha } \right)}}{{\sin \left( {5\alpha } \right)}} & = \mathop {\lim }\limits_{\alpha \to \,0} \left[ {\frac{{12\alpha \sin \left( {12\alpha } \right)}}{{12\alpha }}\frac{{5\alpha }}{{5\alpha \sin \left( {5\alpha } \right)}}} \right] = \mathop {\lim }\limits_{\alpha \to \,0} \left[ {\frac{{12\alpha }}{{5\alpha }}\frac{{\sin \left( {12\alpha } \right)}}{{12\alpha }}\frac{{5\alpha }}{{\sin \left( {5\alpha } \right)}}} \right]\\ & = \mathop {\lim }\limits_{\alpha \to \,0} \left[ {\frac{{12}}{5}\frac{{\sin \left( {12\alpha } \right)}}{{12\alpha }}\frac{{5\alpha }}{{\sin \left( {5\alpha } \right)}}} \right] = \frac{{12}}{5}\left[ {\mathop {\lim }\limits_{\alpha \to \,0} \frac{{\sin \left( {12\alpha } \right)}}{{12\alpha }}} \right]\left[ {\mathop {\lim }\limits_{\alpha \to \,0} \frac{{5\alpha }}{{\sin \left( {5\alpha } \right)}}} \right]\\ & = \frac{{12}}{5}\left( 1 \right)\left( 1 \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{12}}{5}}}\end{align*}