For $x \in R$,$\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x - 3}}{{x + 2}}} \right)^x}$ is equal to

Consider the following statements:
Statement $1$: $\lim _{x \rightarrow 1} \frac{a x^{2}+b x+c}{c x^{2}+b x+a} = 1$ (where $a+b+c \neq 0$).
Statement $2$: $\lim _{x \rightarrow -2} \frac{\frac{1}{x}+\frac{1}{2}}{x+2} = \frac{1}{4}$.

$\mathop {Lim}\limits_{x \to \infty } \frac{2 + 2x + \sin 2x}{(2x + \sin 2x)e^{\sin x}}$ is :

Let $f(x) = 5 - |x - 2|$ and $g(x) = |x + 1|$,where $x \in R$. If $f(x)$ attains its maximum value at $\alpha$ and $g(x)$ attains its minimum value at $\beta$,then $\lim_{x \to \alpha \beta} \frac{(x - 1)(x^2 - 5x + 6)}{x^2 - 6x + 8}$ is equal to:

The limit of $\left[\frac{1}{x^{2}}+\frac{(2013)^{x}}{e^{x}-1}-\frac{1}{e^{x}-1}\right]$ as $x \rightarrow 0$ is:

The value of $\lim _{x \rightarrow \infty}\left(\frac{x^{2}-2 x+1}{x^{2}-4 x+2}\right)^{x}$ is