I found an oblique asymptote of $f(x) = x\left|\frac{x}{x+1}\right|$, but wolfram didn't. Who is correct? Let $f(x) = \frac{x|x|}{|x+1|} =  \frac{x|x|}{|x||1+\frac1x|} =   \frac{x}{|1+\frac1x|}   $

My solution
We are looking for a line $y= \alpha x + \beta$

*

*$ \alpha = \lim_{x \to \pm \infty} \frac{f(x)}{x} = \lim_{x \to \pm \infty} \frac{x}{|1+\frac1x|}  \cdot \frac1x = \lim_{x \to \pm \infty} \frac{1}{|1+\frac1x|} = 1$
Hence, $\boxed{ \alpha =1 }$


*$\beta = \lim_{x \to \pm \infty} f(x) - \alpha x \stackrel{let \text{ } x\to +\infty}{=} \lim_{x \to +\infty} \frac{x}{1 + \frac1x}  - x =\lim_{x \to \pm \infty} \lim_{x \to +\infty} \frac{x^2}{x+1}  - x \\ \quad=\lim_{x \to +\infty} \frac{x^2-x^2-x}{x+1} = - \lim_{x \to \pm \infty}1-\frac1{x+1} = -1$ Hence, $\boxed{\beta = -1}$
Therefore an oblique asymptote is: $\boxed{y =x-1}$
And it seems I am right:


Wolfram's solution

*

*I seached for f(x) asymptotes and it only found a vertical (I did too)

*Then I asked it explicitly for a horizontal and it found none

*Similar query with the word oblique in the comments.


That made me really wonder if there is something wrong with my solution. I triple checked everything and it seems fine. Is there something wrong with my solution? Is $y = x-1$ an asymptote?
 A: As
$$\lim_{x\to\pm\infty}\frac x{x+1}=1,$$ for large $|x|$ you needn't worry about the absolute value. Then it is clear that $f(x)\sim x$, and
$$\lim_{x\to\pm\infty}\left(x\frac x{x+1}-x\right)=\lim_{x\to\pm\infty}-\frac x{x+1}=-1.$$

I suspect that Alpha does not find the asymptote because it is computing in the complex numbers.
A: Here is a faster way for both sides, with asymptotic analysis:
First, we may suppose $x<-1$ or $x>1$ (i.e. $|x>1|$). With this hypothesis, abserve that $\frac x{x+1}>$, so we may remove the absolute value:
$$f(x)=x\,\frac1{1+\frac 1x}=x\biggl(1-\frac1x+\frac1{x^2}+o\Bigl(\frac1{x^2}\Bigr)\biggr)= x-1+\frac1x+o\Bigl(\frac1{x}\Bigr)$$
which proves the line $\:y=x-1\:$ is indeed an asymptote to the curve, and adds that the curve is above its asymptote near $+\infty$ and below its asymptote near $-\infty$.
A: Your second limit evaluation is a little bit confusing with sign but it seems fine
$$\lim_{x \to \pm \infty} x\left|\frac{x}{x+1}\right|-x=\lim_{x \to \pm \infty} x\frac{x}{x+1}-x=\lim_{x \to \pm \infty} \frac{x^2-x^2-x}{x+1}=$$
$$=\lim_{x \to \pm \infty} \frac{-x}{x+1}=\lim_{x \to \pm \infty} \frac{-1}{1+\frac1x}=-1$$
and also your result for the oblique asymptote $y=x-1$ is correct.
