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The limit of the sequence $\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, \dots$ is

  1. $1$
  2. $2$
  3. $2\sqrt{2}$
  4. $\infty$
in Calculus by Boss (17.6k points)
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$2$ is the answer.
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should be 2
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