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

Question

3536 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-06-27T19:22:05

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

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]}