Horizontal Asymptote of Inverse Tangent Function

[Petrus]

Hello MHB,
I got one question, I am currently working with an old exam and I am suposed to draw it with vertican/horizontal lines (and those that are oblique).
\(\displaystyle f(x)=\frac{x}{2}+\tan^{-1}(\frac{1}{x})\)
for the horizontel line
\(\displaystyle \lim_{x->\infty^{\pm}}\frac{x}{2}+\tan^{-1}(\frac{x}{2})\)
Is it enough just to see that \(\displaystyle \lim_{x->\infty^{\pm}}\frac{x}{2} = \pm \infty\) and say there is no horizontel line?
So I have to check oblique line

Regards,
\(\displaystyle |\pi\rangle\)

 

[Ackbach]

Re: Trigometri,limit

Hello MHB,
I got one question, I am currently working with an old exam and I am suposed to draw it with vertican/horizontal lines (and those that are oblique).
\(\displaystyle f(x)=\frac{x}{2}+\tan^{-1}(\frac{1}{x})\)
for the horizontel line
\(\displaystyle \lim_{x->\infty^{\pm}}\frac{x}{2}+\tan^{-1}(\frac{x}{2})\)
Is it enough just to see that \(\displaystyle \lim_{x->\infty^{\pm}}\frac{x}{2} = \pm \infty\) and say there is no horizontel line?

No, it's not. You must also show that the $\tan^{-1}$ term is finite (which it is). Otherwise, you might get an $\infty- \infty$ situation that requires more analysis.

So I have to check oblique line

Regards,
\(\displaystyle |\pi\rangle\)

 

[Petrus]

Re: Trigometri,limit

No, it's not. You must also show that the $\tan^{-1}$ term is finite (which it is). Otherwise, you might get an $\infty- \infty$ situation that requires more analysis.

Thanks for the fast responed!:) Now I know that I should not try think like that!:)

Regards,
\(\displaystyle |\pi\rangle\)