The function $\mathrm{f}$ defined on $$\left(-\frac{1}{3}, \frac{1}{3}\right)$$ by $$\mathrm{f}(x)=\left\{\begin{array}{cc} \frac{1}{x} \log \left(\frac{1+3 x}{1-2 x}\right) & , \quad x \neq 0 \\ \mathrm{k} & , \quad x=0 \end{array}\right.$$ is continuous at $$x=0$$, then $$\mathrm{k}$$ is

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

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