If $f(x) = \begin{cases} x, & x > 1 \\ x^2, & x < 1 \end{cases}$,then $\lim_{x \to 1} f(x) = $

If $f(x) = \operatorname{sgn} \left( 3\cos x - \frac{a}{3} \right)$ is continuous for all $x$,then the least positive integral value of $a$ is - (where $\operatorname{sgn}(x)$ denotes the signum function of $x$)

Discuss the continuity of the function $f$ given by
$f(x) = \begin{cases} x, & \text{if } x \ge 0 \\ x^2, & \text{if } x < 0 \end{cases}$

Let $f(x) = \begin{cases} \frac{x - 4}{|x - 4|} + a, & x < 4 \\ a + b, & x = 4 \\ \frac{x - 4}{|x - 4|} + b, & x > 4 \end{cases}$. Then $f(x)$ is continuous at $x = 4$ when

Discuss the continuity of the function $f$,where $f$ is defined by $f(x) = \begin{cases} -2, & \text{if } x \le -1 \\ 2x, & \text{if } -1 < x \le 1 \\ 2, & \text{if } x > 1 \end{cases}$. Is it continuous at $x=3$?

The function $f(x)=\sqrt{\frac{3 x^2-5 x-2}{2 x^2-7 x+5}}$ has discontinuous points at $x=$