I found this in an asymptote example question but it only has an answer. $$f(x)=x-\sqrt{x^2+5}$$ I solved the $x\to+\infty$ where $\displaystyle\lim_{x\to+\infty}{x-\sqrt{x^2+5}}=\frac{-5}{2}$ so that it has $y=\dfrac{-5}{2}$ for a horizontal asymptote.
It says here that when $x\to-\infty$, it has an oblique asymptote.
How do I find the oblique asymptote of this function when $x\to-\infty$?
On
When x approaches negative infinity, the original function is approximately $f(x)=x-|x|=2x$, so the oblique asymptote is $y=2x$.
When $x$ approaches positive infinity, $f(x)$ should approach 0, leading to a horizontal asymptote of $y=0$.
You can check the result by graphing the function.
On
When $x<0$$$f(x)=x-\sqrt{x^2+5}=x+x\sqrt{1+\frac 5 {x^2}}$$ Now, using equivalents $$f(x)\sim x+x\left(1+\frac 5{2x^2} \right)=2x+\frac 5{2x}$$ which shows the asymptote and how it is approached.
On
(see figure 1 below) I propose here, beyond the good answers you have had from @Claude Leibovici and Michael Rozenberg, a graphical understanding of the issue.
In fact, the (blue) curve associated to function
$$f(x)=y=x-\sqrt{x^2+5}$$
is a branch of hyperbola, explaining the presence of two different asymptotes, thus explaining why the behaviour at $+\infty$ and $-\infty$ is not the same.
The other branch (in red) is associated with the conjugate function :
$$g(x)=y:=x+\sqrt{x^2+5}$$
(minus sign replace by plus sign).
Remark: It is possible to encompass both $f$ and $g$ into a single implicit equation in the following way:
$$\tag{1}y-x=\pm\sqrt{x^2+5} \ \ \iff \ \ (y-x)^2=x^2+5 \ \ \iff \ \ y(y-2x)=5$$
(which is the equation of a hyperbola because, up to an affine change of variables, one has an expression $YX=$ const.).
One could wonder about the interest of equation (1). In fact, it gives the asymptotes ! Here is how: it suffices to set the RHS to $0$ : (1) becomes $x(y-2x)=0$ thus $x=0$ or $y=2x$, the equations of the two asymptotes ! This non orthodox way (transforming a $5$ into a $0$...) can find an explanation, graphical too : consider that you have a family of curves with equations
$$\tag{2}y(y-2x)=z$$ where $z$ is a constant.
and make $z \to 0$... the curve get closer and closer to the asymptotes... as seen on Fig. 2 where $z$ is also considered as a third coordinate.
Figure 1.
Figure 2. The different hyperbolas as level curves of surface $z=x(y-2x)$. "Our" hyperbola (see last equation in relationship (1)) is the upper one, for $z=5$. The smallest is $z$, the closest we get to the asymptotes.
$$\lim_{x\rightarrow+\infty}f(x)=\lim_{x\rightarrow+\infty}\frac{-5}{x+\sqrt{x^2+5}}=0,$$ which says that $y=0$ is a horizontal asymptote for $x\rightarrow+\infty$.
Now, $\lim\limits_{x\rightarrow-\infty}f(x)=-\infty$ and $$\lim_{x\rightarrow-\infty}\frac{f(x)}{x}=1+\lim_{x\rightarrow-\infty}\sqrt{1+\frac{5}{x^2}}=2$$ and it's enough to calculate $\lim\limits_{x\rightarrow-\infty}(f(x)-2x).$
Indeed, $$\lim\limits_{x\rightarrow-\infty}(f(x)-2x)=\lim\limits_{x\rightarrow-\infty}(-x-\sqrt{x^2+5})=\lim_{x\rightarrow+\infty}\frac{-5}{-x-\sqrt{x^2+5}}=0,$$ which says that $y=2x$ is an asymptote of $f$ for $x\rightarrow-\infty$.