If $f(x) = \begin{cases} \frac{x - |x|}{x}, & x \ne 0 \\ 2, & x = 0 \end{cases}$,then

Prove that every rational function is continuous at every point in its domain.

If $f(x) = \begin{cases} \frac{x^2 - 4x + 3}{x^2 - 1}, & x \ne 1 \\ 2, & x = 1 \end{cases}$,then:

If the function $f: R \rightarrow R$ defined by $f(x) = \begin{cases} \frac{a(1-\cos 2x)}{x^2}, & x < 0 \\ b, & x = 0 \\ \frac{\sqrt{x}}{\sqrt{4+\sqrt{x}}-2}, & x > 0 \end{cases}$ is continuous at $x = 0$, then $a+b=$

If $f(x) = \frac{\log_e(1 + x^2 \tan x)}{\sin x^3}, x \neq 0$ is to be continuous at $x = 0$,then $f(0)$ must be equal to

Consider the function $f:(0,2) \rightarrow R$ defined by $f(x)=\frac{x}{2}+\frac{2}{x}$ and the function $g(x)$ defined by $g(x)=\begin{cases} \min \{f(t) : 0 < t \leq x\}, & 0 < x \leq 1 \\ \frac{3}{2}+x, & 1 < x < 2 \end{cases}$. Then,