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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
recategorized | 12 views
0
$2$ is the answer.
0

should be 2
by (347 points)