The Equation

Return

$$|x|=\lim_{h \to \infty}h\sqrt{2(1-\cos(\frac{x}{h}))}$$

Using the Double angle formula for cosine

$$\cos(2\theta)=\sin(\theta)^2-\cos(\theta)^2$$

Pythagoras for trig

$$\sin(\theta)^2+\cos(\theta)^2=1$$

$$\cos(\theta)^2=1-\sin(\theta)^2$$

Substituting it in the Double angle formula

$$\cos(2\theta)=\sin(\theta)^2-(1-\sin(\theta)^2)$$

$$=\sin(\theta)^2+\sin(\theta)^2-1$$

$$=2\sin(\theta)^2-1$$

$$2\theta=\frac{x}{h}$$

$$\cos(\frac{x}{h})=2\sin(\frac{x}{2h})^2-1$$

Substituting into the first equation

$$\lim_{h \to \infty}h\sqrt{2(1-2\sin(\frac{x}{2h})^2-1)}$$

$$=\lim_{h \to \infty}h\sqrt{2(2\sin(\frac{x}{2h})^2)}$$

$$=\lim_{h \to \infty}h\sqrt{2^2\sin(\frac{x}{2h})^2}$$

$$=\lim_{h \to \infty}h\sqrt{(2\sin(\frac{x}{2h}))^2}$$

$$=\lim_{h \to \infty}h\sqrt{(2\sin(\frac{x}{2h}))^2}$$

$$=\lim_{h \to \infty}h|2\sin(\frac{x}{2h})|$$

As $h \to \infty$, $\frac{x}{2h} \to 0$. As $\theta \to 0$, $\sin(\theta) = \theta$

$$=\lim_{h \to \infty}h|2\frac{x}{2h}|$$

$$=\lim_{h \to \infty}h|\frac{x}{h}|$$

$$=\lim_{h \to \infty}|x|$$

$$=|x|$$

$$|x|=\lim_{h \to \infty}h\sqrt{2(1-\cos(\frac{x}{h}))}$$