What is the oblique limit of a function with a hard limit to infinity?

[loreberto911]

Hi everybody, I have this function to study

$\frac{(x+1)}{arctan(x+1)}$

I need the limit to infinity,it's oblique and I have to find q,from y=mx+q.
so

q=lim(x->inf) $\frac{(x+1)}{arctan(x+1)} -2x/\pi$

I don't know how to solve it.the limit gives infinity to me.but calculators online give
$q=\frac{2(2+\pi)}{\pi^2}$

I can't use laurent series
thanks!

Ok I post everything I know, the function is
$f(x)=\frac{(x+1)}{arctan(x+1)}$
I have to do a normal study function, the limit the function to +- inf is inf so I have to find the oblique limit.
the formula is
lim x->inf f(x)-mx-q=0

m=lim x->+-inf f(x)/x=lim $f(x)=\frac{(x+1)}{xarctan(x+1)}$=$\frac{2}{\pi}$

q= lim x->+-inf f(x)-mx = $f(x)=\frac{(x+1)}{arctan(x+1)}$ - $\frac{2x}{\pi}$

=$\frac{\pi x+\pi-2x atan(x+1)}{atan(x+1)\pi}$ =inf

but q has to be $q=\frac{2(2+\pi)}{\pi^2}$

 

[mfb]

I don't know how to solve it.the limit gives infinity to me.

Please show your work then, otherwise it is impossible to tell what went wrong.

I moved the thread to our homework section.

 

[vela]

$$q = \lim_{x \to \infty} [f(x)-mx] = \lim_{x \to \infty} \left[\frac{(x+1)}{\arctan(x+1)}-\frac{2x}{\pi}\right]$$

First, do a change variables to $u=x+1$ to simplify the fraction a bit.
$$q = \lim_{u \to \infty} \left[\frac{u}{\arctan u}-\frac{2(u-1)}{\pi}\right]$$ Then try writing the denominator as $\frac{\pi}{2} + \left(\arctan u - \frac{\pi}{2}\right)$. Can you see where to go from there?