Let $$f(x)=\left\{\begin{array}{cl}\frac{1}{|x|}, & \text { for }|x|>1 \\ a x^2+b, & \text { for }|x| \leq 1\end{array}\right.$$. If $$\lim _\limits{x \rightarrow 1^{+}} f(x)$$ and $$\lim _\limits{x \rightarrow 1^{-}} f(x)$$ exist, then the possible values for $$a$$ and $$b$$ are

$$\frac{d}{d x}\left(\lim _{x \rightarrow 2} \frac{1}{y-2}\left(\frac{1}{x}-\frac{1}{x+y-2}\right)\right)=$$

If $$f(x)=\left\{\begin{array}{cc}\frac{x^2 \log (\cos x)}{\log (1+x)} & , \quad x \neq 0 \\ 0 & , x=0\end{array}\right.$$, then at $$x=0, f(x)$$ is

Let $$f: R^{+} \longrightarrow R^{+}$$ be a function satisfying $$f(x)-x=\lambda$$ (constant), $$\forall x \in R^{+}$$ and $$f(x f(y))=f(x y)+x, \forall x, y, \in R^{+}$$. Then, $$\lim _\limits{x \rightarrow 0} \frac{(f(x))^{1 / 3}-1}{(f(x))^{1 / 2}-1}=$$