Evaluate the given limit

by RoRi

September 3, 2015

Compute

$\lim_{x \to 0} \frac{1 - \cos (2x)}{x^2}.$

Here, we recall the trig identity, $\cos (2x) = 1 - 2 \sin^2 x$ and compute,

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

Stumbling Robot

Evaluate the given limit

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