If the function defined by $f(x) = \begin{cases} \frac{2^x - 2^{-x}}{x}, & x \neq 0 \\ k, & x = 0 \end{cases}$ is continuous at $x = 0$, then $e^k$ is equal to:

  • A
    $\log \left(\frac{2}{e}\right)$
  • B
    $\log 4$
  • C
    $4$
  • D
    $1$

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