0
$\begingroup$

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 $x\to-\infty$?

$\endgroup$
  • $\begingroup$ Double-check your algebra for the limit as $x\to\infty$. By algebra, $f(x)\approx -\dfrac5{2x}$ for $x$ large. $\endgroup$ – Ted Shifrin Dec 30 '17 at 6:22
  • $\begingroup$ @Ted I must've misread. Thank you. $\endgroup$ – Melon Dec 30 '17 at 6:46
0
$\begingroup$

$$\lim_{x\rightarrow+\infty}f(x)=\lim_{x\rightarrow+\infty}\frac{-5}{x+\sqrt{x^2+5}}=0,$$ 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 $x\rightarrow-\infty$.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

(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.

enter image description here

Figure 1.

enter image description here

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.

$\endgroup$
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.