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

Question

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

Isabella Ackerman · Accepted Answer

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# Consequently, #lim_(xrarroo) arctan (x^2)/sqrtx = 0#

Christopher Allers · Answer

undefined To use the Squeeze Theorem to find the limit of Arctan(n^2)/sqrt(n) as n approaches infinity, we need to find two other functions that "squeeze" the given function and have the same limit as n approaches infinity.

First, we can observe that -1 < Arctan(n^2)/sqrt(n) < 1 for all positive values of n. This is because the range of the arctangent function is between -π/2 and π/2, and the denominator, sqrt(n), grows faster than the numerator, n^2, as n approaches infinity.

Next, we can find two functions, f(n) and g(n), such that f(n) ≤ Arctan(n^2)/sqrt(n) ≤ g(n) for all positive values of n. One possible choice is f(n) = -1 and g(n) = 1.

Now, we can take the limit of f(n), Arctan(n^2)/sqrt(n), and g(n) as n approaches infinity. Since f(n) and g(n) both have a limit of 1, by the Squeeze Theorem, the given function Arctan(n^2)/sqrt(n) also has a limit of 1 as n approaches infinity.