Find lim_{x to infinity} {pi/2 - arctan (x)}^1/x ?

1 Answer
Jun 27, 2018

lim_(x->oo) (pi/2-arctanx)/x = 0

Explanation:

Let y= pi/2-arctanx. When x->+oo, y->0 and using the trigonometric identity:

cot(pi/2-theta) = tan theta

we have:

cot y = cot(pi/2-arctanx) = tan(arctanx) = x

Then:

lim_(x->oo) (pi/2-arctanx)/x = lim_(y->0) y/coty

lim_(x->oo) (pi/2-arctanx)/x = lim_(y->0) ytany = 0

graph{(pi/2-arctanx)/x [-10, 10, -5, 5]}