If $$f(a)=2, f^{\prime}(a)=1, g(a)=-1, g^{\prime}(a)=2$$, then as $$x$$ approaches a, $$\frac{\mathrm{g}(x) \mathrm{f}(\mathrm{a})-\mathrm{g}(\mathrm{a}) \mathrm{f}(x)}{(x-\mathrm{a})}$$ approaches

Let $$\mathrm{f}(x)=5-|x-2|$$ and $$\mathrm{g}(x)=|x+1|, x \in \mathrm{R}$$ If $$\mathrm{f}(x)$$ attains maximum value at $$\alpha$$ and $$\mathrm{g}(x)$$ attains minimum value at $$\beta$$, then $$\lim _\limits{x \rightarrow-\alpha \beta} \frac{(x-1)\left(x^2-5 x+6\right)}{x^2-6 x+8}$$ is equal to

$$\lim _\limits{x \rightarrow \infty} x^3\left\{\sqrt{x^2+\sqrt{1+x^4}}-x \sqrt{2}\right\}=$$

$$\lim _\limits{x \rightarrow 0} \frac{\sqrt{1+x \sin x}-\sqrt{\cos x}}{\tan ^2 \frac{x}{2}}=$$