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

  • A
    $-1$
  • B
    $0$
  • C
    $1$
  • D
    $2$

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