Tomamos como el límite
$$\lim_{x \to 0^+}\left(\frac{1 - \cos{\left(x \right)}}{5 x^{2}}\right)$$
Usamos la fórmula trigonométrica
sin(a)^2 = (1 - cos(2*a))/2
cambiamos
$$\lim_{x \to 0^+}\left(\frac{1 - \cos{\left(x \right)}}{5 x^{2}}\right) = \lim_{x \to 0^+}\left(\frac{1 - \cos{\left(x \right)}}{5 x^{2}}\right)$$
=
$$\frac{2 \left(\lim_{x \to 0^+}\left(\frac{\sin{\left(\frac{x}{2} \right)}}{x}\right)\right)^{2}}{5}$$
$$\lim_{x \to 0^+}\left(\frac{\sin{\left(\frac{x}{2} \right)}}{x}\right) = \lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{2 u}\right)$$
=
$$\frac{\lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)}{2}$$
El límite
$$\lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)$$
hay el primer límite, es igual a 1.
entonces
$$\frac{2 \left(\lim_{x \to 0^+}\left(\frac{\sin{\left(\frac{x}{2} \right)}}{x}\right)\right)^{2}}{5} = \frac{2 \left(\frac{\lim_{u \to 0^+}\left(\frac{\sin{\left(u \right)}}{u}\right)}{2}\right)^{2}}{5}$$
=
$$\frac{2}{4 \cdot 5}$$
=
$$\frac{1}{10}$$
Entonces la respuesta definitiva es:
$$\lim_{x \to 0^+}\left(\frac{1 - \cos{\left(x \right)}}{5 x^{2}}\right) = \frac{1}{10}$$