What's the limit of xArctan(x/x^2+1) as x approaches positive infinity ?

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Answer 1 · 2017-07-25T23:04:01

#lim_(x->oo) x arctan(x/(x^2+1)) = 1#

The limit:

#lim_(x->oo) x arctan(x/(x^2+1))#

is in the indeterminate form #+oo xx 0#. We can reconduce it to the form #0/0# and apply l'Hospital's rule:

#lim_(x->oo) x arctan(x/(x^2+1)) = lim_(x->oo) arctan(x/(x^2+1))/(1/x)#

#lim_(x->oo) x arctan(x/(x^2+1)) = lim_(x->oo) (d/dx arctan(x/(x^2+1)))/(d/dx (1/x))#

Now evaluate:

#d/dx arctan(x/(x^2+1)) = 1/(1+ (x/(x^2+1))^2) d/dx (x/(x^2+1))#

#d/dx arctan(x/(x^2+1)) = (x^2+1)^2/(x+ (x^2+1))^2( (x^2+1)-2x^2)/(x^2+1)^2#

#d/dx arctan(x/(x^2+1)) =-(x^2-1)/(x+ (x^2+1))^2#

#d/dx arctan(x/(x^2+1)) =-(x^2-1)/(x^4+2x^2+ x+ 1)#

Then:

#lim_(x->oo) x arctan(x/(x^2+1)) = lim_(x->oo) (-(x^2-1)/(x^4+2x^2+ x+ 1))/(-1/x^2)#

#lim_(x->oo) x arctan(x/(x^2+1)) = lim_(x->oo) x^2((x^2-1)/(x^4+2x^2+ x+ 1))#

#lim_(x->oo) x arctan(x/(x^2+1)) = lim_(x->oo) ((x^4-x^2)/(x^4+2x^2+ x+ 1))#

#lim_(x->oo) x arctan(x/(x^2+1)) = 1#

graph{x arctan(x/(x^2+1)) [-10, 10, -5, 5]}