Jtdylktuy

I found this in an asymptote example question but it only has an answer.
$$f(x)=x-sqrt{x^2+5}$$
I solved the $xto+infty$ where $displaystylelim_{xto+infty}{x-sqrt{x^2+5}}=frac{-5}{2}$ so that it has $y=dfrac{-5}{2}$ for a horizontal asymptote.

It says here that when $xto-infty$, it has an oblique asymptote.

How do I find the oblique asymptote of this function when $xto-infty$?

$$lim_{xrightarrow+infty}f(x)=lim_{xrightarrow+infty}frac{-5}{x+sqrt{x^2+5}}=0,$$
which says that $y=0$ is a horizontal asymptote for $xrightarrow+infty$.

Now, $limlimits_{xrightarrow-infty}f(x)=-infty$ and $$lim_{xrightarrow-infty}frac{f(x)}{x}=1+lim_{xrightarrow-infty}sqrt{1+frac{5}{x^2}}=2$$ and it's enough to calculate $limlimits_{xrightarrow-infty}(f(x)-2x).$

Indeed, $$limlimits_{xrightarrow-infty}(f(x)-2x)=limlimits_{xrightarrow-infty}(-x-sqrt{x^2+5})=lim_{xrightarrow+infty}frac{-5}{-x-sqrt{x^2+5}}=0,$$
which says that $y=2x$ is an asymptote of $f$ for $xrightarrow-infty$.

When $x<0$$$f(x)=x-sqrt{x^2+5}=x+xsqrt{1+frac 5 {x^2}}$$ Now, using equivalents
$$f(x)sim x+xleft(1+frac 5{2x^2} right)=2x+frac 5{2x}$$ which shows the asymptote and how it is approached.

(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:

$$tag{1}y-x=pmsqrt{x^2+5} iff (y-x)^2=x^2+5 iff y(y-2x)=5$$

(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.

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.