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So I am asked to find this limit

$$\lim_{x\to 0} \; \frac{2\cos(a+x)-\cos(a+2x)-\cos(a)}{x^2}$$

I know I am supped to use the trig identity $\cos(u+v)=\cos(u)\cos(v)-\sin(u)\sin(v)$ but I am having trouble with the denominator. I am trying to use the common limit $\lim_{x \to 0} \frac{sin(x)}{x}$ but if I expand out every term, I don't have a sin everywhere. How would I deal with that? I tried simplifying the answer but I am honesty getting nowhere.

Could I have some help with how I would simplify? Thanks.

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  • $\begingroup$ Can you use L'Hospital ? $(-2\sin(a+x)+2\sin(a+2x))/2x$, then $(-2\cos(a+x)+4\cos(a+2x))/2$. $\endgroup$ – Yves Daoust Oct 11 '16 at 13:39
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HINT:

Using Prosthaphaeresis Formula, $$\cos(a+2x)+\cos a=2\cos(a+x)\cos x$$

Now $\dfrac{1-\cos x}{x^2}=\left(\dfrac{\sin x}x\right)^2\cdot\dfrac1{1+\cos x}$

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  • $\begingroup$ Thank you very much. I managed to get the correct answer. I just had to derive your identity on the second line myself but I figured it out too. $\endgroup$ – Future Math person Oct 11 '16 at 13:48
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Use the identity $cos(A) - cos(B) = 2sin(\frac{A+B}{2})sin(\frac{B-A}{2})$

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Hint:

$$L=-\lim_{h\to0}\frac{\dfrac{\cos(a+2h)-\cos(a+h)}{h}-\dfrac{\cos(a+h)-\cos(a)}{h}}{h}.$$

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