Question

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

Answer 1

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

Explanation:

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]}