Using definition of limits, need to find $g'(x)$ for $g(x)= \dfrac{\sin x}{15}$
$g'(x) = \lim_\limits{h \to 0} \dfrac{g(c+h) -g(c)}{h} = \lim_\limits{h \to 0} \dfrac{\dfrac{\sin (x+h) - \sin x}{15}}{h} = \lim\limits_{h \to 0} \dfrac{ \sin x(\cos h-1) + \sin h \cos x}{15h}$
Since $\;\lim_\limits{h \to 0} \dfrac{ \cos h - 1}{h} = 0$
$\lim_\limits{h \to 0} \sin x ( \sin h \cos x)$
From this, I cannot see how I can get to the final answer of $\;\dfrac{\cos x}{15}\;.\;$ I am doing this without the LHopital rule. Where did I go wrong?
$\begin{align} g'(x)&=\lim_\limits{y\to0}\dfrac{g(c+h)-g(c)}h\\[1em] &=\lim_\limits{h\to 0}\dfrac{\frac{\sin(x+h)-\sin x}{15}}h\\[1em] &=\lim_\limits{h\to0}\dfrac{\sin x(\cos h-1)+\sin h\cos x}{15h}\\[1em] &=\lim_\limits{h\to0}\dfrac{\sin x(\cos h-1)}{15h}+\lim_\limits{h\to0}\dfrac{\sin h \cos x}{15h}\\[1em] \end{align}$
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After applying $\lim_\limits{h\to 0}\frac{\cos h-1}h=0$ , we're left with $\;\lim_\limits{h\to0}\dfrac{\sin h\cos x}{15h}\;$ where you can use $\;\lim_\limits{h\to0}\dfrac{\sin h}h=1\;$ to get the answer.