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

Question

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 $$ \lim_{a \to \pm \infty}f'(a)(x-a) = 0 \tag{3}\qquad ? $$

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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?

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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?

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

Answer 1

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

Answer 2

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 $.

Answer 3

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.