Rationalise the denominator by multiplying by $\dfrac{\sqrt{2}+\sqrt{1+\cos x}}{\sqrt{2}+\sqrt{1+\cos x}}$:

Denominator becomes $2-(1+\cos x)=1-\cos x$.

$\lim_{x\to0}\dfrac{\sin^2x(\sqrt{2}+\sqrt{1+\cos x})}{1-\cos x}$

Use $\dfrac{\sin^2x}{1-\cos x}=\dfrac{1-\cos^2x}{1-\cos x}=1+\cos x$:

$=\lim_{x\to0}(1+\cos x)(\sqrt{2}+\sqrt{1+\cos x})=(1+1)(\sqrt{2}+\sqrt{2})=2\cdot2\sqrt{2}=\mathbf{4\sqrt{2}}$

Wait: $2\times2\sqrt{2}=4\sqrt{2}$... but option A is 4. Let me recheck: $(2)(\sqrt{2}+\sqrt{2})=2\cdot2\sqrt{2}=4\sqrt{2}$. Answer: $4\sqrt{2}$ is not listed as option A (which is 4). The correct answer is $\mathbf{4\sqrt{2}}$ (index 1).