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I was trying to find the right oblique asymptote of the following function: $$ g(x)= \frac{x^2+(x+2)\cosh(x)}{\sinh(x)}=\frac{x^2}{\sinh(x)}+(x+2)\coth (x)$$ Now since $\frac{x^2}{\sinh(x)}\to 0$ and $\coth(x)\to 1$ as $x\to \infty$, it is easy to see that this asymptote is $y=x+2$. However, when I try to find this asymptote using L'Hôpitals rule, I get a different result: $$\begin{align} \lim_{x\to \infty}g(x) & =\lim_{x\to \infty} \frac{x^2+(x+2)\cosh(x)}{\sinh(x)} \\ & \stackrel{LH}{=}\lim_{x\to \infty}\frac{2x+\cosh(x)+(x+2)\sinh (x)}{\cosh(x)} \\ & =\lim_{x\to \infty} \frac{2x}{\cosh(x)}+1+(x+2)\tanh(x) \\ & =\lim_{x\to \infty} x+3 \end{align}$$ since $\frac{2x}{\cosh(x)}\to 0$ and $\tanh(x)\to 1$ as $x\to \infty$. This suggests that the asymptote is $y=x+3$ instead of $y=x+2$.

A quick look at the function using WolframAlpha shows that $y=x+2$ is indeed the correct asymptote, so I highly suspect that I somehow applied L'Hôpitals rule in a wrong way. I have however no clue as to what I did wrong. Could anyone enlighten me?

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  • $\begingroup$ Please note that that’s limit is completely meaningless for the calculation of the asymptote and also the way to calculate with de l’Hopital is uncorrect. $\endgroup$
    – user
    Dec 10, 2017 at 1:20
  • $\begingroup$ Yeah okay. So I guess the problem with this approach is that it is completely arbitrary in what parts of the limit are evaluated and what parts are left as a linear approximation? $\endgroup$
    – Grimp0w
    Dec 10, 2017 at 1:25
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    $\begingroup$ If $f$ is $any$ function such that $\lim_{x\to \infty}f(x)=\infty$ then $\infty=$ $\lim_{x\to\infty} f(x)=$ $\lim_{x\to\infty}x=$ $=\lim_{x\to \infty}x+1=$ $\lim_{x\to \infty}x+2=$ $\lim_{x\to \infty}x+3.$ But that tells us nothing about asymptotes. We have $\lim_{x\to \infty}g(x)-(x+2)=0,$ which is why the line $y=x+2$ is an asymptote of the curve $y=g(x).$ $\endgroup$
    – DanielWainfleet
    Dec 10, 2017 at 1:26
  • $\begingroup$ Exactly you can’t calculate the limit in this way throwing away some parts and keeping the others. You have to calculate it properly and as described here below. $\endgroup$
    – user
    Dec 10, 2017 at 1:28

2 Answers 2

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You seem to be assuming that L'Hôpital preserves asymptotes, when it's not the case.

Take for instance $$ \frac{x^2(x+2)}{x^2}, $$ with the obvious asymptote $x+2$. If you take derivatives to use L'Hôpital, you get $$ \frac{3x^2+4x}{2x}=\frac{3x+4}{2}, $$ and the asymptote is not the same.

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    $\begingroup$ The problem here is not that l’Hopital do not preserve the asymptote, the problem is that the limit is completely meaningless to calculate the asymptote. $\endgroup$
    – user
    Dec 10, 2017 at 1:16
  • $\begingroup$ @Martin Argerami Ah okay, I missed that. Thanks! Do you also happen to know why asymptotes aren't preserved? $\endgroup$
    – Grimp0w
    Dec 10, 2017 at 1:16
  • $\begingroup$ I cannot give a deep reason. L'Hôpital addresses the fact that $f(x)/g(x)$ and $f'(x)/g'(x)$ have the same limit at infinity if the latter exists, but it doesn't tell you anything about the two quotients behaving the same. $\endgroup$
    – Martin Argerami
    Dec 10, 2017 at 1:25
  • $\begingroup$ The wikipedia article on L'Hôpital says something about its geometric interpretation here. Does this imply that the slope of the asymptote is preserved? $\endgroup$
    – Grimp0w
    Dec 10, 2017 at 1:34
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    $\begingroup$ @Grimp0w: Read the wikipedia article more carefully. It talks about a parametric curve tracing $(f(t),g(t))$, and so if $f'(t):g'(t)$ tends to a constant ratio, then $f(t):g(t)$ also. It says nothing about a curve defined as the graph of a function of one coordinate, in this case tracing $(f(x)/g(x),x)$. $\endgroup$
    – user21820
    Dec 10, 2017 at 7:25
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To find the asymptote:

$$y=mx+n$$

you should calculate separately the following limits:

$$m=\frac{g(x)}{x}$$

for the slope, and

$$n=g(x)-mx$$

for the intercept.

Take also a look here:

How to find the oblique asymptote of root of a function?

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