How do you use the Squeeze Theorem to find $lim Arctan(n^2)/sqrt(n)$ as x approaches infinity?

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Answer 1 · 2015-10-18T01:19:32

See the explanation, below.

For all $x$ , we know that $x^2 > 0$ , so we have

$0 <= arctan (x^2) < pi/2$ .

For positive $x$ , $sqrtx > 0$ , so we can divide the inequalitiy without changing the directions of the inequalities.

$0 <= arctan (x^2)/sqrtx < pi/(2sqrtx)$ .

$lim_(xrarroo)0 = 0$ and $lim_(xrarroo) pi/(2sqrtx) = 0$

Therefore,

$lim_(xrarroo) arctan (x^2)/sqrtx = 0$

How do you use the Squeeze Theorem to find lim Arctan(n^2)/sqrt(n)lim Arctan(n^2)/sqrt(n) as x approaches infinity?