$$\lim_{h\to 0}\frac{\sin(x+h)-\sin(x)}{h}$$ Using the fact that: $$\lim_{x\to 0}\frac{\sin(x)}{x}=1.$$
Thank you in advance for any help.
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2Start using the identity for the sine of the sum of two angles. – user1337 May 03 '15 at 13:15
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Simply expand $\sin (x+h)$ and you will find. – Yes May 03 '15 at 13:16
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1Depending on how you do it, you might also need $$\lim_{x\to 0} \dfrac {\cos(x) - 1}{x} = 0$$ – May 03 '15 at 13:20
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Use $\sin(x+h)=\sin(x)\cos(h)+\sin(h)\cos(x)$ for starting. – Claude Leibovici May 03 '15 at 13:21
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http://www.coloringsun.com/wp-content/uploads/2014/09/A-Girl-with-an-Open-Book-Coloring-Page.jpg – John Joy May 03 '15 at 15:52
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As told in one of the comments use the following identity:-$$\sin(x+h)=\sin(x)\cos(h)+\sin(h)\cos(x)$$ Now plugging this in you get$$\lim_{h\to 0}\frac{\sin(x+h)-\sin(x)}{h}= \lim_{h\to 0}\frac{\sin(x)\cos(h)+\sin(h)\cos(x)-\sin(x)}{h}$$ this becomes $$\lim_{h\to 0}\frac{\sin(x)[\cos(h)-1]+{\sin(h)\cos(x)}}{h}$$ now dividing the h between the two numerator expression this becomes $$\lim_{h\to 0}[{\frac{{\sin(x)[\cos(h)-1]}}{h}+\frac{{\sin(h)\cos(x)}}{h}}]$$ here as the limit $\lim_{h\to 0}{\frac{{\cos(h)-1}}{h}}$ goes to zero, we're left with $$\cos(x)\lim_{h\to0}\frac{\sin(h)}{h}$$ and using the idea you mentioned this becomes $\cos(x)$
Chris
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where you say "this becomes", I think you want to say $\lim_{h\to 0}\frac{\sin(x)[\cos(\color{red}h)-1]+{\sin(h)\cos(x)}}{h}$ – Lucas May 03 '15 at 16:02