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$$\lim_{h\to 0}\frac{\cos(x_0+h)-\cos(x_0)}h \quad\text{as } x_0\in(0,\pi)$$

I did actually do it without L'Hopital rule as I just multiplied the top and bottom of the conjugate of the top of the fraction and just went from there, using the addition formula for the cosine. But this was extremely tedious. I was wondering if there is an easier way.

The answer to this gives $-\sin(x_0)$.

snowman
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    If $x_0\in(0,\pi)$ and $x_0\ne\pi/2$ (that is to say $\cos x_0\ne 0$) then you should get $\infty$ as an answer, not $-\sin x_0$ – Mher Nov 26 '14 at 13:35
  • sorry I missed out part of the question. I edited it now. – snowman Nov 26 '14 at 13:37

3 Answers3

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Using difference formula for cosines, we get $$\frac{\cos(x_0+h)-\cos(x_0)}{h}= \frac{2\sin(-h/2)\sin(x_0+\frac{h}{2})}{h}=$$ $$=-\sin\left(x_0+\frac{h}{2}\right)\cdot \frac{\sin(h/2)}{h/2}\to -\sin(x_0)$$ as $h\to 0$, since $$\lim_{h\to 0}\frac{\sin h}{h} = 1$$

Mher
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$$\lim_{h\to 0}\frac {\cos(x_0+h)-\cos (x_0)}{h}=\left.\frac {d}{dx}\cos (x)\right|_{x=x_0}=-\sin (x_0)$$

Dario
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$\cos(x_0+h)=\cos x_0\cos h-\sin x_0\sin h$, so your limit can be rewritten as $$ \lim_{h\to0}\left(\cos x_0\frac{\cos h-1}{h}-\sin x_0\frac{\sin h}{h}\right) $$ What's the limit of $\frac{\sin h}{h}$ is known; you should also know $$ \lim_{h\to0}\frac{\cos h-1}{h}= \lim_{h\to0}\frac{\cos^2h-1}{h(\cos h+1)}= \lim_{h\to0}\frac{\sin h}{h}\frac{-\sin h}{\cos h+1}=\dots $$

egreg
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