# What is asymptote to `oo` if y=arctg(x/(x+1))?

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You need to evaluate the equation of slant asymptote `y = mx + n` , such that:

`m = lim_(x->oo)((f(x))/x)`

Let ` y = f(x),` such that:

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

`m = (arctan 1)/oo = (pi/4)/oo -> 0`

Since `m != +- oo` , you may evaluate n, such that:

`n =lim_(x->oo) (f(x) - mx) => n = lim_(x->oo) f(x)`

`n = lim_(x->oo) arctan(x/(x + 1)) => n = arctan lim_(x->oo) (x/(x + 1)) => n = arctan 1 => n = pi/4`

**Hence, evaluating the equation of the slant asymptote to the graph of the function, yields `y = pi/4` , thus, since m = 0, the slant asymptote becomes a horizontal asymptote.**

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