Question
Using the \lim\limits_{x\to 0} \dfrac{\sin x}{x}=1 and factor identity to prove derivative of sine: (\sin x)^{'} = \cos x
Answer
\lim\limits_{h\to 0} \dfrac{\sin (x+h)-\sin x}{h}
= \lim\limits_{h\to 0} \dfrac{2\sin \dfrac{h}{2} \cos (x+\dfrac{h}{2} )}{h}
= \lim\limits_{h\to 0} \dfrac{\sin \dfrac{h}{2}}{ \dfrac{h}{2}} \cos (x+\dfrac{h}{2} )
= \lim\limits_{h\to 0} \dfrac{\sin \dfrac{h}{2}}{ \dfrac{h}{2}} \cdotp \lim\limits_{h\to 0} \cos (x+\dfrac{h}{2} )
=\cos x