slant asymptote of $y=x\operatorname{arccot}(x)$ help me with this problem

Find the slant asymptote of

$$y=x\operatorname{arccot}(x)$$

I know that the slant asymptote is $y=ax+b$ if

$$\lim_{x\to \infty} f(x)-(ax+b)=0$$

the answer given to me is:
$$\lim_{x\to +\infty} \frac{y}{x}=\lim_{x\to +\infty}\operatorname{arccot}(x)=0$$
$$\lim_{x\to -\infty} \frac{y}{x}=\lim_{x\to -\infty}\operatorname{arccot}(x)=\pi$$
$\lim_{x\to -\infty}(y-\pi x)=\lim_{x\to -\infty}\frac{\operatorname{arccot}(x)-\pi}{\frac{1}{x}}=\lim_{x\to -\infty}\frac{-\frac{1}{1+x^2}}{-\frac{1}{x^2}}=1$(L'Hospital)

conclusion: $y=\pi x +1$ is the slant asymptote of the function

i don't understand this answer can someone explain the answer for me or have any other ways to do this problem?

note: this is from the last year exam of my college so i don't think the question is wrong, the answer was made by students though so they could be wrong.
Note that,

$$\lim_{x\to \infty} x \cot^{-1}(x) = \lim_{x\to \infty} \frac{ \tan^{-1}\frac1x}{\frac1x}=1$$

Thus, the asymptote $y=b=1$ is horizontal, not slant.

Edit:

The range of the inverse cotangent function is $(-\frac\pi2,0),(0,\frac\pi2)$, which does not permit a slant asymptote as $x\rightarrow -\infty$. Some source may specify its range as $(0,\pi)$, as in your answer. Then, a slant asymptote could be identified, which may be questionable.
Note that,

$$\lim_{x\to \infty} x \cot^{-1}(x) = \lim_{x\to \infty} \frac{ \tan^{-1}\frac1x}{\frac1x}=1$$

Thus, the asymptote $y=b=1$ is horizontal, not slant.

Edit:

The range of the inverse cotangent function is $(-\frac\pi2,0),(0,\frac\pi2)$, which does not permit a slant asymptote as $x\rightarrow -\infty$. Some source may specify its range as $(0,\pi)$, as in your answer. Then, a slant asymptote could be identified, which may be questionable.