If $$f(x)=\left\{\begin{array}{cc}x^2-1, & 0< x<2 \\ 2 x+3, & 2 \leq x<3\end{array}\right.$$,

the quadratic equation whose roots are $$\lim _\limits{x \rightarrow 2^{-}} f(x)$$ and $$\lim _\limits{x \rightarrow 2^{+}} f(x)$$ is

$$\lim _\limits{y \rightarrow 0} \frac{\sqrt{3+y^3}-\sqrt{3}}{y^3}=$$

Consider the following statements

Statement 1 : $$\lim _\limits{x \rightarrow 1} \frac{a x^2+b x+c}{x^2+b x+a}$$ is 1

(where $$a+b+c \neq 0$$).

Statement 2 : $$\lim _\limits{x \rightarrow -2} \frac{\frac{1}{x}+\frac{1}{2}}{x+2}$$ is $$\frac{1}{4}$$.

If $$f(x)=\left|\begin{array}{ccc}\cos x & 1 & 0 \\ 0 & 2 \cos x & 3 \\ 0 & 1 & 2 \cos x\end{array}\right|$$, then $$\lim _\limits{x \rightarrow \pi} f(x)$$ is equal to