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Find the limit:

$$\lim_{h\rightarrow0}\frac{f(x+h)−f(x)}{h}$$ Given that $f(x)=\sin(2x)$. Tried many ways, but I kept on getting an indeterminate form. I can't find a way to cancel out terms on the numerator and denominator.

Any help will be appreciated.

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    $\begingroup$ The limit is the deriavative $f'(x)$ which is $2 \cos (2x)$. $\endgroup$ – Kavi Rama Murthy Jan 25 at 0:07
  • $\begingroup$ $\dfrac{f(x+h)-f(x)}{h}$ is called the difference quotient $\endgroup$ – gen-z ready to perish Jan 25 at 1:54
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Hint

$$\sin(2x+2h)=\sin 2x\cos 2h+\cos 2x\sin 2h$$ $$\lim_{u\to 0}{\sin u\over u}=1$$ $$\lim_{u\to 0}{\cos u-1\over u}=0$$

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Use http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html

$$\sin2a(x+h)-\sin2ax=2\sin(ah)\cos(2ax+ah)$$

and $\lim_{y\to0}\dfrac{\sin y}y=1$

Here $a=1$

Similarly for $\cos$ if required

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