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#