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I am trying to prove the following limit. $$ \lim_{\Delta\theta\rightarrow0+}\frac{\sqrt{2-2\cos\Delta\theta}}{\Delta\theta}=1 $$ I am going to use this result to prove derivatives of sine and cosine, so I am looking for a proof that won't compute this one by using L'Hôpital's rule. Maybe there is a δ,ε-proof or something else?

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  • $\begingroup$ I do not think you are going to be able to find a $\delta$-$\epsilon$ proof. Compare this: math.stackexchange.com/questions/75130/… $\endgroup$
    – Miguel
    Commented Aug 29, 2020 at 20:01
  • $\begingroup$ MathJax works in the title section too. $\endgroup$
    – Shaun
    Commented Aug 29, 2020 at 20:08

4 Answers 4

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First, $\displaystyle 2-2\cos x = 4 \sin^2 \Big( \frac x2 \Big),$ and so, for $x >0$, $$\sqrt{2-2\cos x} = 2\, \Big|\sin \Big( \frac x2 \Big)\Big| = 2 \sin \Big( \frac x2 \Big).$$ Now your limit simply becomes: $$\lim_{x \to 0^+} \dfrac{\sin \Big( \dfrac x2 \Big)}{\dfrac x2} = \lim_{t \to 0^+} \frac{\sin t}{t}.$$

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$$\lim\limits_{x \to 0+}\frac{\sqrt{1-\cos x}}{x}= \lim\limits_{x \to 0+}\frac{\sqrt{2\sin^2 \frac{x}{2}}}{x}= \lim\limits_{x \to 0+}\frac{\sqrt{2}\sin \frac{x}{2}}{x}=\frac{1}{\sqrt{2}} $$

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  • $\begingroup$ For those who want to edit: I especially have not "$2$" under square root to emphasize trigonometry identity used, I move it out of limit. Multiplied my answer on $\sqrt{2}$ gives exactly answer $1$. $\endgroup$
    – zkutch
    Commented Aug 29, 2020 at 21:57
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By standard limit, we have that

$$\frac{\sqrt{2-2\cos\Delta\theta}}{\Delta\theta}=\sqrt 2\sqrt{\frac{1-\cos\Delta\theta}{\Delta\theta^2}} \to \sqrt 2\,\sqrt{\frac12}=1$$

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Let $\varepsilon>0$. Let $0<x-0=x<\delta$ and choose $\varepsilon =\delta$. We know that $2-2\cos(x)=4\sin^2(\frac{x}{2})$ and that $\sin(\frac{x}{2})\leq \frac{x}{2}$ for $x>0$. $$0\leq\bigg|\frac{2\sin(\frac{x}{2})}{x}-1\bigg|\leq\bigg|\frac{2\cdot\frac{x}{2}-x}{x}\bigg|=0<\delta=\varepsilon\ $$

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