What is the y-limit of the inverse tangent function?

[LtIvan]

Allo,
When I was experimenting with graphing functions, I noticed the inverse tangent, or arctanget, curves away from y=2, or may be less. What is the y limit for the inverse tangent function? Does it for ever increase, or terminate at a co-ordinate?

 

[S.G. Janssens]

What is the y limit for the inverse tangent function? Does it for ever increase, or terminate at a co-ordinate?

First graph the function $\tan$ on the real line. Is it defined everywhere? Then recall that $\arctan$ is defined as the inverse of the restriction of $\tan$ to the interval $(-\tfrac{\pi}{2},\tfrac{\pi}{2})$. (Why is it necessary to first restrict $\tan$?) Finally try to answer your own question.

 

[member 587159]

First graph the function $\tan$ on the real line. Is it defined everywhere? Then recall that $\arctan$ is defined as the inverse of the restriction of $\tan$ to the interval $[-\tfrac{\pi}{2},\tfrac{\pi}{2}]$. (Why is it necessary to first restrict $\tan$?) Finally try to answer your own question.

Little correction: you should exclude pi/2 and -pi/2 from the interval as tan is not defined for those values.

 

[LtIvan]

I approximately understand. So by your logic, when you graph y=tan(x), when it goes up and approaches endlessly, it stops round about π/2 and goes back down to increase and repeat this process? My original question was when does it stop on the y axis? It is infinity? However, when I apply this in my calculator.
tan(π/2)≈0.027 (2sf)
This does not make sense; when I graph this using a software. The line does not define this? when I observe the x=π/2 is does not intersect tan at 0.027? I am missing something?

Thanks in advance,

 

[member 587159]

I approximately understand. So by your logic, when you graph y=tan(x), when it goes up and approaches endlessly, it stops round about π/2 and goes back down to increase and repeat this process? My original question was when does it stop on the y axis? It is infinity? However, when I apply this in my calculator.
tan(π/2)≈0.027 (2sf)
This does not make sense; when I graph this using a software. The line does not define this? when I observe the x=π/2 is does not intersect tan at 0.027? I am missing something?

Thanks in advance,

You should put your calculator ib radian mode, not degree mode, if you do not have done this yet. tan(pi/2) must give an error, otherwise you or your calculator are doing something wrong.

 

[LtIvan]

Okay thanks, that worked. Thinking immaterially, correct me if I am invalid, when π/2 subtract an infinitesimal, tan function would equal what one would interpret as infinity?
Let i="infinitesimal"
tan(π/2-i)=∞
?

However, thanks for your help, I now understand things better.

 

[member 587159]

Okay thanks, that worked. Thinking immaterially, correct me if I am invalid, when π/2 subtract an infinitesimal, tan function would equal what one would interpret as infinity?
Let i="infinitesimal"
tan(π/2-i)=∞
?

However, thanks for your help, I now understand things better.

I don't like to talk about infinitesimals, however:

lim x-> + infinity arctan(x) = pi/2

 

[chiro]

Hey LtIvan,

You should define the y-limit mathematically and evaluate it using limit laws.

If something increases forever then the derivative is greater than zero.

Evaluating functions are easy if you define them consistently, correctly, and concisely.