If $f(x)=\begin{cases} \frac{x-3}{|x-3|}+a & , x<3 \\ a+b & , x=3 \\ \frac{|x-3|}{x-3}+b & , x>3 \end{cases}$ is continuous at $x=3$,then the value of $a-b$ is

If a function $f(x)$ defined by $f(x) = \begin{cases} ax^2 + bx + c, & x \leq -1 \\ 2x^2 + 4x + 1, & -1 < x < 1 \\ cx^2 + bx + a, & x \geq 1 \end{cases}$ is continuous on $\mathbb{R}$,and $\lim_{x \rightarrow \frac{3}{2}} f(x) = 14$,then find $\lim_{x \rightarrow -2} f(x)$.

At $x=0$, the function $f(x) = \begin{cases} \frac{x}{|x|+2x^2}, & x \neq 0 \\ k, & x=0 \end{cases}$ is:

If the function $f(x) = \frac{\sqrt{1+x}-1}{x}$ is continuous at $x=0$,then $f(0) = $

$A$ function $y=f(x)$ with $f(-1)=-249$ has no maximum and has only one minimum at $x=5$ with $f(5)=75$. Which one of the following is true?

Is the function defined by $f(x) = \begin{cases} x + 5, & \text{if } x \le 1 \\ x - 5, & \text{if } x > 1 \end{cases}$ a continuous function?