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Establish the given limit formula

Prove the formula for the limit:

    \[ \lim_{x \to 0} \frac{1- \cos x}{x^2} = \frac{1}{2}. \]

Using,

    \[ \lim_{x \to 0} \frac{\sin x}{x} = 1. \]


Proof. We multiply the numerator and denominator by 1 + \cos x,

    \[ \lim_{x \to 0} \frac{1-\cos x}{x^2} = \lim_{x \to 0} \frac{1 - \cos^2 x}{x^2 (1+\cos x)}. \]

Then, by the Pythagorean identity, \cos^2 x + \sin^2 x = 1 \implies 1 - \cos^2 x = \sin^2 x, so

    \begin{align*}  \lim_{x \to 0} \frac{1 - \cos^2 x}{x^2 (1+\cos x)} &= \lim_{x \to 0} \left(\frac{\sin^2 x}{x^2} \cdot \frac{1}{1+\cos x} \right) \\  &= \lim_{x \to 0} \left( \left( \frac{\sin x}{x}\right)^2 \left(\frac{1}{1+\cos x} \right) \right) \\  &= \lim_{x \to 0} \left( \left(\frac{\sin x}{x} \right) \left( \frac{\sin x}{x} \right) \left( \frac{1}{1+\cos x} \right) \right) \\  &= \lim_{x \to 0} \frac{1}{1 + \cos x} \\  &= \frac{1}{2}. \qquad \blacksquare \end{align*}

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