Calculer les limites suivantes :

\[ \begin{array}{ll} \text{1) } \displaystyle \lim_{x \to 0} \frac{\sin 3x}{7x} & \text{7) } \displaystyle \lim_{x \to 0} \frac{\sqrt{1 + \sin x} - \sqrt{1 - \sin x}}{x} \\ & \\ \text{2) } \displaystyle \lim_{x \to 0} \frac{\tan 5x}{\sin x} & \text{8) } \displaystyle \lim_{x \to 0} \frac{1 - \cos^3 x}{x \cdot \sin 4x} \\ & \\ \text{3) } \displaystyle \lim_{x \to 0} \frac{1 - \cos 2x}{x^2} & \text{9) } \displaystyle \lim_{x \to 0} \frac{1 - \cos 5x}{(x^2 + x)\tan 3x} \\ & \\ \text{4) } \displaystyle \lim_{x \to 0^+} \frac{\tan x - \sin x}{\sqrt{x}} & \text{10) } \displaystyle \lim_{x \to 1} \frac{1 - x^2}{\sin \pi x} \\ & \\ \text{5) } \displaystyle \lim_{x \to 0} \frac{\sin x - \tan x}{x^3} & \text{11) } \displaystyle \lim_{x \to \frac{\pi}{4}} \frac{\tan^2 x + 2\tan x - 3}{\sin x - \cos x} \\ & \\ \text{6) } \displaystyle \lim_{x \to 0} \frac{\sin x \cdot \sin 2x}{1 - \cos x} & \text{12) } \displaystyle \lim_{x \to \frac{\pi}{3}} \frac{\cos 2x + \cos x}{2\cos x - 1} \end{array} \]