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Find the limit:

$$\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$

Given that $f(x) = \cos(2x)$

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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The OP posted a comment below nbubis's answer asking what to do with $\frac{\sin^2(\frac{h}{2})}{h}$. I can't comment back, (no reputation at all) so here's my contribution, multiply $\frac{h}{4}$ in the numerator and denominator. This will lead to $\frac{\sin^2(\frac{h}{2})}{(\frac{h}{2})^2}*\frac{h}{4}$. The first term is just one squared, the second term is $0$. So the term evaluates to zero. Accept his answer.

EDIT: I decided to post my own version of the answer anyway.

We know that, $\cos(C)-\cos(D)=2\sin(\frac{C+D}{2})\sin(\frac{D-C}{2})$

Now, Let $C=2x+h, D=2x$

Hence, our limit becomes, $\frac{2\sin(\frac{4x+h}{2})sin(\frac{-h}{2})}{\frac{h}{2}}$

Now, $\frac{\sin(\frac{-h}{2})}{h}$ evaluates to $-\frac{1}{2}$. Proceed.

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  • $\begingroup$ I'd +1 his answer if I could, but again rep. :p $\endgroup$ – Guy Feb 25 '14 at 4:26
  • $\begingroup$ I finally got it, thank you so much! $\endgroup$ – Shaikat Haque Feb 25 '14 at 4:51
  • $\begingroup$ You will always get help on math.se $\endgroup$ – Guy Feb 25 '14 at 4:59
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You want to compute $$ \lim_{h\to0}\frac{\cos (2x+2h)-\cos 2x}{h} $$ But $$ \begin{align*} \cos (2x+2h)-\cos 2x&=\cos 2x\cos 2h-\sin 2x\sin 2h-\cos2x\\ &=(\cos 2x\cos2h-\cos2x)-\sin2x\sin2h\\ &=\cos2x(\cos2h-1)-\sin2x\sin2h\\ &=\cos2x(1-2\sin^2h-1)-\sin2x\sin2h\\ &=-2\cos2x\sin^2h-\sin2x\sin2h \end{align*} $$ Thus the above limit can be written as $$ -2\cos2x\lim_{h\to0}\left(\frac{\sin h}{h}\sin h\right)-\sin2x\lim_{h\to0}\frac{\sin2h}{h} $$ Using the following limit $$ \lim_{\alpha\to0}\frac{\sin\alpha}{\alpha}=1 $$ we have: $$ \begin{align*} \lim_{h\to0}\left(\frac{\sin h}{h}\sin h\right)&=\lim_{h\to0}\frac{\sin h}{h}\times\lim_{h\to0}\sin h\\ &=1\times0\\ &=0 \end{align*} $$ and $$ \begin{align*} \lim_{h\to0}\frac{\sin2h}{h}&=2\lim_{2h\to0}\frac{\sin2h}{2h}\\ &=2 \end{align*} $$ Hence $$ \lim_{h\to0}\frac{\cos (2x+2h)-\cos 2x}{h}=-2\sin2x $$

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Hint: $$f(x+h)-f(x)=\cos(2x)\cos(h)-\sin(2x)\sin(h)-\cos(2x)$$ $$=\cos(2x)(1-\cos h)-\sin(2x)\sin(h)$$ $$=2\cos(2x)\sin^2(h/2)-\sin(2x)\sin(h)$$

Now use what you know about the limit $\sin (h)/h$.

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  • $\begingroup$ I know sin(h)/h will become 1. But what can I do about (sin^2(h/2) )/h? $\endgroup$ – Shaikat Haque Feb 25 '14 at 4:13
  • $\begingroup$ We are not allowed to use L'hopital's rule $\endgroup$ – Shaikat Haque Feb 25 '14 at 4:16
  • $\begingroup$ @ShaikatHaque - $\sin(h/2)/h$ is like writing $2\sin(x)/x$ do you see why? Now you have an extra $\sin(h)$ which just goes to zero. $\endgroup$ – nbubis Feb 25 '14 at 4:22
  • $\begingroup$ It's clear now, thank you for your help! $\endgroup$ – Shaikat Haque Feb 25 '14 at 4:52
  • $\begingroup$ @nbubis wouldn't it be 1/2 sin x/x? $\endgroup$ – Flowers Feb 25 '14 at 6:58
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As an additional way, you can use the chain rule. $$\lim\limits_{h\to 0}\frac{\cos(2x+2h)-\cos(2x)}{h}=\lim\limits_{h\to 0}\frac{(2x+2h)-2x}{(2x+2h)-2x}\frac{\cos(2x+2h)-\cos(2x)}{h}=\\\lim\limits_{h\to 0}\frac{\cos(2x+2h)-\cos(2x)}{(2x+2h)-2x}\frac{(2x+2h)-2x}{h}$$Let u = 2x and g = 2h $$=\lim\limits_{h\to 0}\frac{\cos(u+g)-\cos(u)}{g}\cdot2 = \lim\limits_{g\to 0}\frac{\cos(u+g)-\cos(u)}{g}\cdot2 = \\2\lim\limits_{g\to 0}\frac{\cos u\cos g - \sin u \sin g -\cos u}{g} = 2\left (\lim\limits_{g\to 0}\frac{\cos u(\cos g - 1)}{g} - \sin u \right) = 2\left (\lim\limits_{g\to 0}\cos u\frac{\cos^2 g - 1}{g(\cos g + 1)} - \sin u \right)= 2\left (\lim\limits_{g\to 0}\cos u\frac{\sin g}{\cos g + 1} - \sin u \right)=-2\sin u = -2 \sin(2x)$$

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