If $f(x) = \begin{cases} x, & \text{when } 0 \le x \le 1 \\ 2 - x, & \text{when } 1 < x \le 2 \end{cases}$,then $\lim_{x \to 1} f(x) = $

If $f(x) = \cos x$ when $x = n\pi$ $(n = 0, 1, 2, 3, \dots)$ and $f(x) = 3$ otherwise,and $\phi(x) = \begin{cases} x^2 + 1 & \text{when } x \neq 3, x \neq 0 \\ 3 & \text{when } x = 0 \\ 5 & \text{when } x = 3 \end{cases}$,then find $\lim_{x \to 0} f(\phi(x))$.

Let $S$ be the set of all $(\alpha, \beta) \in \mathbb{R} \times \mathbb{R}$ such that $\lim _{x \rightarrow \infty} \frac{\sin(x^2)(\log_e x)^\alpha \sin(1/x^2)}{x^{\alpha \beta}(\log_e(1+x))^\beta} = 0$. Then which of the following is (are) correct?

$\lim _{x \rightarrow \frac{\pi}{4}} \frac{2 \sqrt{2}-(\cos x+\sin x)^3}{1-\sin 2 x}=$

The value of $\mathop {\lim }\limits_{x \to \infty } \frac{(x + 1)(3x + 4)}{x^2(x - 8)}$ is equal to

If ${x_n} = \frac{{1 - 2 + 3 - 4 + 5 - 6 + \dots - 2n}}{{\sqrt {{n^2} + 1} + \sqrt {4{n^2} - 1} }},$ then $\mathop {\lim }\limits_{n \to \infty } {x_n}$ is equal to