Ta thấy $\sin 3x+1 \ge 0 \forall x\Rightarrow \left|\sin 3x+1 \right|=\sin 3x+1$$\Rightarrow \lim_{x\to0} f(x)=\frac{\left|1-(1+\sin 3x)\right|}{\sqrt{2\sin^2\dfrac x2}}$
$=\lim_{x\to0}\left|\frac{\sin 3x}{\sqrt 2\sin \dfrac x2} \right|=\lim_{x\to 0}\left|\frac{6}{\sqrt2}.\frac{\sin 3x}{3x}.\frac{\dfrac x2}{\sin \dfrac x2} \right|$
$=\frac 6{\sqrt 2}=3\sqrt 2$