How to find the limit of $\frac{1−\cos 5x}{x^2}$ as $x\to 0$? What is the right approach to calculate the Limit of $(1-\cos(5x))/x^2$ as $x \rightarrow 0$?
From Wolfram Alpha, I found that:
$$\lim_{x \to 0} \frac{1- \cos 5x}{x^2} = \frac{25}{2}.$$
How do I get that answer?

$\cos(2\alpha) = 1-2\sin ^2 \alpha$
Therefore
$\frac{1-\cos (5x)}{x^2}=\frac{2 \sin ^2 \left( \frac{5x}{2}\right)}{x^2} = \frac{25}{2} \left( \frac{\sin \frac{5}{2}x}{\frac{5}{2}x} \right)^2 \rightarrow \frac{25}{2}$
