# A limit at infinity (horizontal asymptote) of function involving the inverse tangent function

#### Petrus

##### Well-known member
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

##### Indicium Physicus
Staff member
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

##### Well-known member
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,
$$\displaystyle |\pi\rangle$$