Limit of an inverse tangent function

[greenteacup]

Homework Statement

limit as x->$$\infty$$ of arctan[(1/x)-1]

2. The attempt at a solution

Well, my graphing calculator is showing a vertical asymptote at x=0 and a horizontal one at x=(what I think is) -$$\pi$$/4

I'm not really sure how to solve this though... hence me being here : /

EDIT: Nevermind, I figured it out by splitting it up into lim(arctan(1/x)) and lim(arctan(1)), which did indeed give me -pi/4. I'm not sure how to delete this though, so I'll just leave it up for anyone who's having a similar problem, I suppose. :)

 

[Dick]

Homework Statement

limit as x->$$\infty$$ of arctan[(1/x)-1]

2. The attempt at a solution

Well, my graphing calculator is showing a vertical asymptote at x=0 and a horizontal one at x=(what I think is) -$$\pi$$/4

I'm not really sure how to solve this though... hence me being here : /

EDIT: Nevermind, I figured it out by splitting it up into lim(arctan(1/x)) and lim(arctan(1)), which did indeed give me -pi/4. I'm not sure how to delete this though, so I'll just leave it up for anyone who's having a similar problem, I suppose. :)

You got the right answer, but you did it wrong. arctan(1/x-1) is not equal to arctan(1/x)-arctan(1). You can't break it up that way. If you know lim(1/x)=0 then lim(1/x-1)=(-1). Then since arctan is continuous, lim(arctan(1/x-1))=arctan(lim(1/x-1))=arctan(-1).

 

[njama]

Ok take y=(1/x)-1. What y goes to as x goes to infinity?