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) = $

Find $\mathop {\lim }\limits_{x \to 1} f(x)$,where $f(x) = \begin{cases} x^{2}-1, & x \leq 1 \\ -x-1, & x > 1 \end{cases}$

$\mathop {\lim }\limits_{x \to 0} \cos \frac{1}{x}$

If $f(x) = \begin{cases} x^2 - 3, & 2 < x < 3 \\ 2x + 5, & 3 < x < 4 \end{cases}$,the equation whose roots are $\lim_{x \to 3^-} f(x)$ and $\lim_{x \to 3^+} f(x)$ is

The value of $\lim _{x}$ ${\rightarrow \frac{\pi}{2}} \frac{\left(1-\tan \left(\frac{x}{2}\right)\right)(1-\sin x)}{\left(1+\tan \left(\frac{x}{2}\right)\right)(\pi-2 x)^3}$ is

$\mathop {\lim }\limits_{x \to \pi /2} \frac{{{a^{\cot x}} - {a^{\cos x}}}}{{\cot x - \cos x}} = $