# Asymptotes of y = x - arctanx

• Aug 2nd 2010, 01:08 PM
SyNtHeSiS
Asymptotes of y = x - arctanx
Sketch the graph of y = x - arctanx, showing its asymptotes.

Now I got the correct shape of the graph, but my asymptotes were wrong.

I dont understand how you would get the asymptote of y = x + π/2 and y = x - π/2. Could someone please explain?
• Aug 2nd 2010, 01:18 PM
Math Major
Remember that tangent is not a one-to-one function (ie, it sends multiple points in its domain to the same point in its range). Thus, in order for arctangent to be a function, its range has to be restricted to $\displaystyle -\frac{\pi}{2} < x < \frac{\pi}{2}$, otherwise it would be multiply defined (without the restriction, where, for instance, would arctan(0) go?). So, to find the asymptotes, we need to find where the graph does not pass.

We know that $\displaystyle |arctan(x)| < \frac{\pi}{2}$, so where are those asymptotes?
• Aug 3rd 2010, 04:03 AM
SyNtHeSiS
Quote:

Originally Posted by Math Major
We know that $\displaystyle |arctan(x)| < \frac{\pi}{2}$, so where are those asymptotes?

I know that arctanx has horizontal asymptotes of y = π/2 and y = -π/2 (inversion of tanx's undefined x values) but I dont see any x values in y = x - arctanx which would make it undefined?
• Aug 3rd 2010, 04:14 AM
HallsofIvy
If you are talking about vertical asymptotes, it has no vertical asymptotes.
• Aug 3rd 2010, 07:10 AM
SyNtHeSiS
Quote:

Originally Posted by HallsofIvy
If you are talking about vertical asymptotes, it has no vertical asymptotes.

Yeah thats what I meant, that it has no vertical/horizontal asymptotes, which is why I wanted to know, how can you tell that y = x - arctanx has oblique asymptotes?
• Aug 3rd 2010, 08:26 AM
skeeter
Quote:

Originally Posted by SyNtHeSiS
Yeah thats what I meant, that it has no vertical/horizontal asymptotes, which is why I wanted to know, how can you tell that y = x - arctanx has oblique asymptotes?

as $\displaystyle x \to \pm \infty$ , $\displaystyle \arctan{x} \to \pm \frac{\pi}{2}$.

so ... as $\displaystyle x \to \pm \infty$ , $\displaystyle y = x - \arctan{x}$ approaches the line $\displaystyle y = x$
• Aug 4th 2010, 05:50 AM
SyNtHeSiS
Quote:

Originally Posted by skeeter
as $\displaystyle x \to \pm \infty$ , $\displaystyle \arctan{x} \to \pm \frac{\pi}{2}$.

so ... as $\displaystyle x \to \pm \infty$ , $\displaystyle y = x - \arctan{x}$ approaches the line $\displaystyle y = x$

Sorry but I am a bit confused. I understand the first line, but for the second line, how does it approach y=x? I thought the lines it approaches were y= x+/- π/2?
• Aug 4th 2010, 07:06 AM
skeeter
Quote:

Originally Posted by SyNtHeSiS
Sorry but I am a bit confused. I understand the first line, but for the second line, how does it approach y=x? I thought the lines it approaches were y= x+/- π/2?

I'll agree with that ...

as $\displaystyle x \to \infty$ , the function approaches $\displaystyle y = x - \frac{\pi}{2}$

as $\displaystyle x \to -\infty$ , the function approaches $\displaystyle y = x + \frac{\pi}{2}$
• Aug 4th 2010, 01:43 PM
SyNtHeSiS
Quote:

Originally Posted by skeeter
I'll agree with that ...

as $\displaystyle x \to \infty$ , the function approaches $\displaystyle y = x - \frac{\pi}{2}$

as $\displaystyle x \to -\infty$ , the function approaches $\displaystyle y = x + \frac{\pi}{2}$

One more question, if you use a limit with x tending to +/- infinity, why isnt it y = ∞ - π/2 and y = ∞ + π/2?
• Aug 4th 2010, 01:50 PM
skeeter
Quote:

Originally Posted by SyNtHeSiS
One more question, if you use a limit with x tending to +/- infinity, why isnt it y = ∞ - π/2 and y = ∞ + π/2?

there is a difference between the limit of a function and the limit of a function value. $\displaystyle \infty \pm \, any \, constant$ is still $\displaystyle \infty$
• Mar 26th 2013, 05:38 AM