If $f(x) = \begin{cases} \frac{\sin 5x \tan kx}{x^2} & , x \neq 0 \\ 1 & , x = 0 \end{cases}$ is continuous at $x = 0$,then the value of $k$ is . . . . . . . $(\because k \neq 0)$

  • A
    $1/5$
  • B
    $1/3$
  • C
    $1/15$
  • D
    $5$

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