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Some references show that to find an oblique asymptote of a function $f(x)$, we must see the limit of $$ m = \lim_{x \rightarrow \pm \infty} \frac{f(x)}{x} $$ If $m \ne 0$ and finite, then there is an oblique asymptote of the form $y = mx + c$. However, I think it would be more intuitive by searching the limit of $$ \lim_{x \rightarrow \pm \infty} f'(x) $$ If this limit exists, then we can determine the asymptote.

Question : Am I correct if I generalize the 2nd one for finding an oblique asymptote?

I have not seen any reference to use the second one (limit of $f'$) for finding an oblique asymptote. But it is more intuituive.., and we can also see from the first one that $\lim \limits_{x \rightarrow \pm \infty} \frac fx $ has an indefinite form $\frac{\infty}{\infty} $, then by L'Hopital it can be equal to $\lim f'(x)$.

Thanks in advance.

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  • $\begingroup$ It is equal to $\lim f'(x)$ is the latter exists, which is not always the case. So your definition would break down in several points, so you would presumably need special cases to make it work. Also, it's not necessarily more intuitive; once you get familiar with the concept of asymptotic equivalence (en.wikipedia.org/wiki/Asymptotic_analysis) , it's clear that the first condition expresses the fact that $f(x)$ grows "more or less" like $x$, i.e. $f(x)$ is close to a line, i.e. $f(x)$ it has an oblique asymptote $\endgroup$ – Ant Oct 24 '17 at 17:06
  • $\begingroup$ @Ant . This is in terms of basic Calculus, for graphing functions. An asymptote line cannot be crossed (at the limit as $x$ grows unbounded )..? $\endgroup$ – Arief Anbiya Oct 24 '17 at 17:10
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    $\begingroup$ @Arief: Your last statement is not correct. There's no Calculus text that I'm aware of which prohibits, in any way, the crossing of a function and its asymptote. In particular the function $$f(x) = \frac{\sin(x)}{x}$$ is asymptotic to $y=0$, as $x$ approaches $\infty$. $\endgroup$ – quasi Oct 24 '17 at 17:15
  • $\begingroup$ Thanks @quasi . If $f(x)$ is not permitted to cross the line, I thought the 2nd method would be ok. But there is a counter-example below. $\endgroup$ – Arief Anbiya Oct 24 '17 at 17:44
  • $\begingroup$ @Arief: Note also: The standard definition applies to functions which are not necessarily differentiable. By the way, good question (I already gave you +1). $\endgroup$ – quasi Oct 24 '17 at 17:51
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For a more straightforward counterexample, take $f(x) = \ln(x)$.

Its derivative $f'(x)=\frac{1}{x}$ limits to $0$ as $x \to +\infty$.

But this function has no horizontal asymptote. In fact, $\lim_{x \to \infty} \ln(x) = +\infty$ so the graph goes arbitrarily far above every horizontal line as $x \to +\infty$.

You can modify this example in many ways. For instance, $f(x) = x + \ln(x)$ has derivative limiting to $1$ as $x \to +\infty$, but the graph of $y=f(x)$ goes arbitrarily far above every slope 1 line as $x \to +\infty$, hence it has no slope 1 asymptote.

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  • $\begingroup$ Thanks. This counter-example is quite interesting, its almost like it has a moving asymptote.. $\endgroup$ – Arief Anbiya Oct 24 '17 at 17:24
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Let $f(x)=x+\sin{\sqrt{x}}.$ Then $f'(x)=1+\dfrac{\cos{\sqrt{x}}}{2\sqrt{x}} \underset{x\to+\infty}{\to} 1,$ but oblique asymptote does not exists since $\nexists \lim\limits_{x\to+\infty}(f(x)-x).$

Another example: The function $g(x) = x+\dfrac{\sin{x^2}}{x}$ has an asymptote $y=x,$ but it's derivative $g'(x)=1+2\cos{x^2}-\dfrac{\sin{x^2}}{x^2}$ does not have a limit as $x\to\infty.$

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  • $\begingroup$ Thanks. In your example, the line $y = x$ will be crossed periodically. But isn't an asymptote line cannot be crossed by the function $f(x)$ (as $x$ grows unbounded)..? This is for basic Calculus in graphing functions. $\endgroup$ – Arief Anbiya Oct 24 '17 at 17:15
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    $\begingroup$ The first example can be modified to $f(x) = x+\sin{\sqrt{x}}-1-\frac{1}{x}.$ But, according to Wikipedia: Some sources include the requirement that the curve may not cross the line infinitely often, but this is unusual for modern authors. $\endgroup$ – M. Strochyk Oct 24 '17 at 19:20
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This works for rational functions, but not for more general functions. For example, $$ \lim_{x \to \infty} \frac{x+\sin{x}}{x} = 1, $$ but the derivative of $x+\sin{x}$ is $1+\cos{x}$, which does not have a limit as $x \to \infty$.

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  • $\begingroup$ Thanks @Chappers. But $f(x) = x + \sin(x)$ does not have an oblique asymptote.. If a function has, than we can first see it by seeing that limit of $f(x)$ is unbounded.. and then check whether or not its gradient will tend to a constant as $x$ grows without limit. $\endgroup$ – Arief Anbiya Oct 24 '17 at 16:56
  • $\begingroup$ The line with gradient $1$ is not an oblique asymptote for $f(x) = x + \sin(x)$.. isn't it? $\endgroup$ – Arief Anbiya Oct 24 '17 at 17:04
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    $\begingroup$ @Arief: Asymptotic refers to closeness, not slope. Two functions $f,g$ are asymptotic as $x$ approaches $\infty$ if and only if $$\lim_{x \to \infty}\,\bigl(f(x)-g(x)\bigr)=0$$ So you're right: $x + \sin(x)$ does not have $y=x$ an oblique asymptote. In fact, it doesn't have an oblique asymptote. $\endgroup$ – quasi Oct 24 '17 at 17:04

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