# $δ,ε$-proof for limit $\frac{\sqrt{2-2\cos(x)}}{x}$.

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?

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

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: $$\lim_{x \to 0^+} \dfrac{\sin \Big( \dfrac x2 \Big)}{\dfrac x2} = \lim_{t \to 0^+} \frac{\sin t}{t}.$$
$$\lim\limits_{x \to 0+}\frac{\sqrt{1-\cos x}}{x}= \lim\limits_{x \to 0+}\frac{\sqrt{2\sin^2 \frac{x}{2}}}{x}= \lim\limits_{x \to 0+}\frac{\sqrt{2}\sin \frac{x}{2}}{x}=\frac{1}{\sqrt{2}}$$
• 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 $1$. Commented Aug 29, 2020 at 21:57
$$\frac{\sqrt{2-2\cos\Delta\theta}}{\Delta\theta}=\sqrt 2\sqrt{\frac{1-\cos\Delta\theta}{\Delta\theta^2}} \to \sqrt 2\,\sqrt{\frac12}=1$$
Let $$\varepsilon>0$$. Let $$0 and choose $$\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\$$