$\mathop {\lim}\limits_{x \to 1} \left[ {\left[ {\frac{4}{{{x^2} - {x^{ - 1}}}} - \frac{{1 - 3x + {x^2}}}{{1 - {x^3}}}} \right]^{ - 1} + \frac{{3 \cdot ({x^4} - 1)}}{{{x^3} - {x^{ - 1}}}}} \right] = $

$\mathop {\lim }\limits_{x \to 0^ + } \frac{x e^{1/x}}{1 + e^{1/x}} = $

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:

$\lim _{x \rightarrow 0} \frac{\sin x \sin ^{-1} x}{x^2}$ is equal to

$\lim _{x \rightarrow \infty}\left(\frac{2 x^2+3 x+4}{x^2-3 x+5}\right)^{\frac{3|x|+1}{2|x|-1}} = $

If $l = \lim_{x \rightarrow 0} \frac{x}{|x| + x^2}$,then the value of $l$ is