$\lim_{x\to 0} \frac{1-\cos(x)}{x\cos(x)} $ without L'hopital Please help me to solve this limit without using L'Hôpital's rule. I don't know what other method can't be used to solve this limit.
$$\lim_{x\to 0} \frac{1-\cos(x)}{x\cos(x)}  $$

Use Taylor series
$$\cos(x) \approx 1 - \frac{x^2}{2}$$ thence
$$\frac{1 - \cos(x)}{x\cos(x)} = \frac{1 - \left(1 - \frac{x^2}{2}\right)}{x\cdot\left(1 - \frac{x^2}{2}\right)} = \frac{x^2}{2x - x^3} = \frac{x^2}{x^2\left(\frac{2}{x} - x\right)} = \frac{1}{\frac{2}{x}} = \frac{x}{2}$$
And since $x\to 0$ the limit is
$$\boxed{0}$$
