Questions mathématiques diverses

Question 2:
Sans utiliser la regle de l'Hopital, calculer la limite suivante
$\lim\limits_{x \to 0}\dfrac{1-\cos x\sqrt{\cos 2x}}{x^2}$

Réponse 2:
$$\begin{align*} &\lim\limits_{x \to 0}\dfrac{1-\cos x\sqrt{\cos 2x}}{x^2}\\ &=\lim\limits_{x \to 0}\dfrac{1-\cos x+\cos x-\cos x\sqrt{\cos 2x}}{x^2}\\ &=\lim\limits_{x \to 0}\dfrac{1-\cos x}{x^2}+\dfrac{(1-\sqrt{\cos 2x})\cos x}{x^2}\\ &=\lim\limits_{x \to 0}\dfrac{1-\cos x}{x^2}+\dfrac{\cos x}{x^2}(1-\sqrt{\cos 2x})\\ &=\lim\limits_{x \to 0}\dfrac{1-\cos x}{x^2}+\dfrac{\cos x}{x^2}\dfrac{1-\cos 2x}{1+\sqrt{\cos 2x}}\\ &=\lim\limits_{x \to 0}\dfrac{1-\cos x}{x^2}+\dfrac{\cos x}{1+\sqrt{\cos 2x}}\dfrac{1-\cos 2x}{x^2}\\ &=\lim\limits_{x \to 0}\dfrac{1-\cos x}{x^2}+\dfrac{\cos x}{1+\sqrt{\cos 2x}}\dfrac{1-\cos 2x}{(2x)^2}\times 4\\ &=\dfrac{1}{2}+\dfrac{1}{2}\times\dfrac{1}{2}\times4\\ &=\dfrac{3}{2} \end{align*}$$