# If $f(x)$ has an asymptote, when does the limit of the tangent lines approach the asymptote?

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 $$$$\lim_{t \to \tau} \frac{\lvert ax(t) + by(t) - c\rvert}{\sqrt{a^2 + b^2}} = 0 \tag{4}$$$$

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!

• You deal with $\lim_{|a|\to\infty}$ but you say nothing about $x$ . – hamam_Abdallah May 4 at 21:35
• @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. – Robert Lee May 4 at 21:52
• You want $\lim_{a\to\pm\inftt}F(a,x)$ , a function of two variables. is $x$ close to $a$ or $x$ is fixed. – hamam_Abdallah May 4 at 21:56
• @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. – Robert Lee May 4 at 22:03
• 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. – Robert Lee May 4 at 22:33

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 $$f(a) - af'(a) \to b$$.

• This is such a clean solution. Once you explained it by rearranging the terms in the line equations the conditions made total sense. +1 – Robert Lee May 4 at 23:14
• +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. – Andrew D. Hwang May 5 at 11:16

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.

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 $$x$$.

• 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. – Robert Lee May 4 at 23:07