$\mathop {\lim }\limits_{x \to 1} f(x)$ is equal to,where $f(x) = \begin{cases} \frac{e^{\frac{1}{x-1}} - 2}{e^{\frac{1}{x-1}} + 2} & x \neq 1 \\ 1 & x = 1 \end{cases}$

  • A
    $-1$
  • B
    $1$
  • C
    $0$
  • D
    does not exist

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