Ta có $\dfrac{(1-\sqrt{\cos x})(e^x -1)}{3\sin^2 x}=\dfrac{(1-\cos x)(e^x -1)}{(1+\sqrt{\cos x}). 12\sin^2 \dfrac{x}{2} \cos^2 \dfrac{x}{2}}$
$=\dfrac{2\sin^2 \dfrac{x}{2}(e^x -1)}{(1+\sqrt{\cos x}). 12\sin^2 \dfrac{x}{2} \cos^2 \dfrac{x}{2}}=\dfrac{e^x -1}{(1+\sqrt{\cos x}). 6\cos^2 \dfrac{x}{2}} =A$
Mà $\lim \limits_{x \to 0} A = 0$