calculus - How to evaluate the limit $lim_{x to infty}left(left(x+frac{1}{x}right)arctan(x)-frac{pi}{2}xright)$?

$$\lim\limits_{x \to \infty}\left(\left(x+\frac{1}{x}\right)\arctan(x)-\frac{\pi}{2}x\right)$$

From graphs, I know that the limit is exactly $-1$ (this function limit is "$b$" from $f(x)=ax+b$, where $f(x)$ is asymptote of another function).

I managed to get to a point, by using l'Hospital rule, where my limit equals 0. I've checked calculations few times, and can't find, where the mistake is.

Could you please show me steps, how to evaluate this limit?

Answer

Since
$$\lim\limits_{x \to \infty}\frac{\arctan(x)}{x}=0$$
and
$$\lim\limits_{x \to \infty}\left(\arctan(x)-\frac{\pi}{2}\right)=0$$

you have
$$\lim\limits_{x \to \infty}\left(\left(x+\frac{1}{x}\right)\arctan(x)-\frac{\pi}{2}x\right)$$
$$=\lim\limits_{x \to \infty}\left(x\arctan(x)-\frac{\pi}{2}x\right)$$
$$=\lim\limits_{x \to \infty}\left(x \left(\arctan(x)-\frac{\pi}{2}\right)\right)$$
$$=\lim\limits_{x \to \infty}\frac{\arctan(x)-\frac{\pi}{2}}{1/x}$$
which has the indeterminate form $0/0.$ Now use l'Hospital
$$=\lim\limits_{x \to \infty}\frac{1/(1+x^2)}{-1/x^2}$$
$$=\lim\limits_{x \to \infty}\frac{-x^2}{1+x^2}=-1$$