Dividiamo numeratore e denominatore con x²: $$\underset{x\to 0}{\lim }\frac{3{\tan }^{2}x}{1-\cos x}=\underset{x\to 0}{\lim }\frac{3\frac{{\tan }^{2}x}{x^2}}{\frac{1-\cos x}{x^2}}$$ dopo qualche manipolazione algebrica si arriva a: $$\underset{x\to 0}{\lim }\frac{3{\tan }^{2}x}{1-\cos x}=\underset{x\to 0}{\lim }\frac{\frac{3}{{\cos }^{2}x}{\left(\frac{\sin x}{x}\right)}^2}{\frac{1-\cos x}{x^2}}=\frac{\underset{x\to 0}{\lim }\frac{3}{{\cos }^{2}x}{\left(\underset{x\to 0}{\lim }\frac{\sin x}{x}\right)}^2}{\underset{x\to 0}{\lim }\frac{1-\cos x}{x^2}}$$ Ricordiamo ora i limiti notevoli: $$\underset{y\to 0}{\lim }\frac{\sin y}{y}=1\underset{y\to 0}{\lim }\frac{1-{\cos }^{2}y}{y}=\frac{1}{2}$$ e calcoliamo il limite: $$\underset{x\to 0}{\lim }\frac{3{\tan }^{2}x}{1-\cos x}=\frac{\underset{x\to 0}{\lim }\frac{3}{{\cos }^{2}x}{\left(\underset{x\to 0}{\lim }\frac{\sin x}{x}\right)}^2}{\underset{x\to 0}{\lim }\frac{1-\cos x}{x^2}}=\frac{3\cdot 1}{\frac{1}{2}}=6$$