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I was looking at functions with horizontal asymptotes. By a basic definition, $f(x)$ has a horizontal asymptote at $y=c$ if $$ \lim_{x \to \pm\infty} f(x) - c= 0 \tag {1} $$ where the $\pm$ indicates one sign or the other, or possibly both. On the other hand, the equation of the tangent line to $f(x)$ at $x=a$ is given by $$ y = f'(a)(x-a) + f(a) \tag{2} $$ So my question is, given that $f$ has a horizontal asymptote at $y=c$, when does $$ \lim_{a \to \pm\infty}f'(a)(x-a) + f(a) = c\qquad ? $$ Or equivalently, since $\lim_{a \to \pm\infty} f(a) =c$ by our hypothesis $(1)$, when does $$ \lim_{a \to \pm \infty}f'(a)(x-a) = 0 \tag{3}\qquad ? $$


While trying to answer this question I ran into a problem with the definition of an asymptote that I've been using. As explained in this answer, by defining an asymptote simply as a function that satisfies the limit you may have functions such as $f(x) = \frac{\sin(x)}{x}$ which (just by following the limit definition) has a horizontal asymptote at $x=0$, but where the continuous limit $\lim_{a \to \infty} f'(a)(x-a)$ is indeterminate since the slope $f'(a)$ will continue to oscillate indefinitely.

What does seem obvious to me is that given some $f(x)$ that's smooth enough, if $f(x)$ approaches the asymptote $y=c$ monotonically then the limit $\lim_{a \to \infty} f'(a)(x-a) = c$. But even though this seems obvious graphically (as the functions starts to "smooth out" into a more linear behavior) I couldn't seem to mathematically describe this, and in turn, I couldn't show that the limit in question holds.

Does anyone have any ideas on how to formally describe the above behavior such that it can be used to prove that limit $(3)$ exists?


Lastly, I wanted to know how I could generalize this definition (and proof of the limit) for oblique asymptotes. Given some curve $\alpha(t)= (x(t), y(t))$, and some line $\ell :\{(x,y) \in \mathbb{R}^2 \vert ax+by -c =0\}$, if we say that $\ell$ is an asymptote to $\alpha$ as $t \to \tau$ for some value of $\tau$ when $$ \lim_{t \to \tau} d(\alpha(t), \ell)=0 $$ where $d(\alpha(t), \ell)$ represents the distance between the curves, then, by recalling the distance from a point to a line, we can generalize equation $(1)$ as \begin{equation} \lim_{t \to \tau} \frac{\lvert ax(t) + by(t) - c\rvert}{\sqrt{a^2 + b^2}} = 0 \tag{4} \end{equation}

So the more general question becomes: If $(4)$ holds, then when does this also imply that $$ \lim_{a \to \tau} \alpha'(a)t + \alpha(a) = \ell \qquad ? $$ And if it doesn't always hold, what other hypothesis does $\alpha(t)$ need to verify such that it does hold?


Any and all help or ideas will be greatly appreciated. Thank you!

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  • $\begingroup$ You deal with $\lim_{|a|\to\infty}$ but you say nothing about $ x $ . $\endgroup$ May 4, 2021 at 21:35
  • $\begingroup$ @hamam_Abdallah, I didn't specify this because in that instance I'm saying that limit of the tangent-line functions $T_a(x)$ happens to tend to the constant function $g(x) = c$. So if this is true, then $$\left(\lim_{a\to \infty}T_a(x)\right)\Big\vert_{x=\zeta} = g(x)\Big\vert_{x=\zeta} = c$$ for whichever value of $x$ I want. $\endgroup$
    – Robert Lee
    May 4, 2021 at 21:52
  • $\begingroup$ You want $\lim_{a\to\pm\inftt}F(a,x) $ , a function of two variables. is $ x $ close to $ a $ or $ x $ is fixed. $\endgroup$ May 4, 2021 at 21:56
  • $\begingroup$ @hamam_Abdallah, since in this particular case I'm claiming that the limit of the tangent lines is a constant, then $x$ can be close to $a$ or fixed or any other value since the whole line will eventually tend to a constant. By fixing $x$ I would essentially be looking at the limit of the tangent lines point-by-point, but since I'm proposing that the whole tangent line tends to a constant line, then whichever value of $x$ you choose to fix should give the same constant limit. If my understanding is wrong please let me know, but I hope I explained why I left it like it is written. $\endgroup$
    – Robert Lee
    May 4, 2021 at 22:03
  • $\begingroup$ I believe this is no longer an issue for the general case I wrote at the end since I'm proposing that the limit of $T_\eta(t) = \alpha'(\eta) t+\alpha(\eta)$ should tend to another function of $t$, namely the line $$\ell(t) = (b,-a)t + \left(0,\frac{c}{b}\right)$$ The thing is that in the horizontal asymptote case we get $a=0$ and $b=1$, so in the above equation the line $\ell(t)$ happens to be $$\ell(t) = \left(t,c\right), \qquad \forall t \in \mathbb{R}$$ where we see that the $y$-coordinate of the function $\ell(t)$ equals $c$ for whichever value of $t$ you want. $\endgroup$
    – Robert Lee
    May 4, 2021 at 22:33

3 Answers 3

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In order to answer this, we need a notion of "limit" for lines. A natural choice in this setting is to parametrize non-vertical lines by slope $m$ and $y$-intercept $b$, so that a line $y = b(a)x + m(a)$ approaches $y = mx + b$ if and only if $m(a) \to m$ and $b(a) \to b$.

As you note, if $f$ is differentiable everywhere, the tangent line to the graph $y = f(x)$ at $x = a$ is $y = f(a) + f'(a)(x - a) = f'(a)x + [f(a) - af'(a)]$. This approaches $y = f(a)$ if and only if $f'(a) \to 0$ and $af'(a) \to 0$. The second clearly implies the first, since $|f'(a)| < |a|\, |f'(a)|$ for $|a| > 1$. It's also fairly clear by example that $f'(a) \to 0$ does not imply $af'(a) \to 0$.

In sum, the tangent line of $f$ at $a$ approaches $y = f(a)$ as $|a| \to \infty$ if and only if $af'(a) \to 0$.

If instead you have an oblique asymptote $y = mx + b$, the differentiable function $g(x) = f(x) - mx$ has a horizontal asymptote, so this question reduces to the previous case, and the necessary and sufficient conditions are $f'(a) \to m$ and $f(a) - af'(a) \to b$.

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  • $\begingroup$ This is such a clean solution. Once you explained it by rearranging the terms in the line equations the conditions made total sense. +1 $\endgroup$
    – Robert Lee
    May 4, 2021 at 23:14
  • $\begingroup$ +1 for a good question. My first thought was "$f'(a) \to 0$". While that's true for rational functions, it's not true for all differentiable functions. $\endgroup$ May 5, 2021 at 11:16
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This works if you i) translate everything to the origin and ii) work in a limited domain, so that $x$ is not allowed to be too far from the point of tangency $a$ and iii) Make some assumptions on $f$. Let's use monotonicity and concavity for example.

Let $f(x)$ be an increasing, concave-down smooth function on $\mathbb{R}$ such that $L = \lim_{x\rightarrow \infty} f(x)$ exists. Define the functions $$T_a(x) = f(a) + f'(a) (x-a).$$ Translate this to the origin: define $S_a(x) = T_a(x+a) = f(a) + f'(a) x$.

Claim: For any $M > 0$, as $a \rightarrow \infty$, $S_a(x) - f(x+a) \rightarrow 0$ uniformly for $x \in [-M,M]$.

Proof: First we prove $\lim_{x\rightarrow \infty} f'(x) = 0$. For suppose not; then there is $\epsilon > 0$ so that for any $L > 0$ we have some $a>L$ with $|f'(a)| > \epsilon$. Since $f(x)$ is assumed increasing, $f'(a) > 0$, so we can say $f'(a) > \epsilon$. Then using the fact that $f'(x)$ is decreasing, we have

\begin{align*} f(x) &= f(a) + f(x) - f(a) \\ &= f(a) + \int_a^x f'(t) \, dt\\ &\geq f(a) + \int_a^x f'(a) \, dt \\ &= f(a) + \epsilon (x-a)\end{align*} which tends to $\infty$ as $x \rightarrow \infty$ contradicting the fact that $\lim_{x\rightarrow \infty} f(x)$ exists and is finite. So $\lim_{x\rightarrow \infty} f'(x) = 0$

Now fix M. Given $\epsilon$ choose $L$ so that for $x>L$ we have both $|f(x)-L| < \epsilon/3$ and $|f'(x)| < \epsilon/(3M)$.

Then for $a > L$ and any $|x| < M$, we have \begin{align*}|S_a(x) - f(x+a)| &= |f(a) - f(x+a) + f'(a) x |\\ & |f(a) - f(x+a)| + |f'(a) x |\\ &|f(a) - L| + |L - f(x+a)| + |f'(a)| x \\&< 2\epsilon/3 + (\epsilon/(3M)) M \\ &= \epsilon.\end{align*}

This is for a horizontal asymptote. Likely you can do a similar argument for an oblique asymptote.

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Take $$f(x)=\frac 1x$$ then $ c=0$ but $$\lim_{a\to\infty}f'(a)(2a-a)=0$$ $$\lim_{a\to\infty}f'(a)(a^2-a)=-1$$ and $$\lim_{a\to\infty}f'(a)(a^3-a)=\infty$$

So, your limit $\lim_{a\to\infty}f'(a)(x-a) $ should depend on $ x $.

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  • $\begingroup$ I'm sorry. I believe I made a poor job at explaining what I meant by "limit of a line". What I mean by taking the limit of the line is somewhat "the same" as when we talk about the convergence of some sequence of functions $f_n(x)\to f(x)$, as we take $n \to \infty$, but in this case the $f_n(x)$ are the tangent lines and the $f(x)$ is the asymptote. Using this same notation, you've showed that $\lim_{n \to \infty} f_n(n^2) \neq \lim_{n \to \infty} f_n(n^3)$, but this is not what I meant in my question. I again apologize for the misunderstanding. $\endgroup$
    – Robert Lee
    May 4, 2021 at 23:07

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