In this post we will investigate the connection between slant asymptotes and the behavior of the derivative at \(\infty\) for differentiable real functions.

We say that a function \(f(x)\) has a slant asymptote \(y=mx +q\) for \(x\to+\infty\) whenever

\begin{equation}\lim_{x\to+\infty} \underbrace{\left[f(x)-mx-q\right]}_{D(x)} = 0,\tag{1}\label{eq2624:1}\end{equation}

i.e. the distance \(D(x)\) between the function and the straight line of equation \(y=mx+q\) tends to \(0\) as \(x \to +\infty\).

In the Figure below a very simple example is depicted, where \(f(x) = \frac1{\sqrt x} + \frac12 x -1\) is the dark blue line and the asymptote \(y = \frac12 x -1\) is the light blue line. Marked in red, you can see the distance between \(f(x)\) and the straight line, that is

\[D(x) = f(x)-mx-q = \frac1{\sqrt x}.\]

Suppose that equation \eqref{eq2624:1} is satisfied. Then we have

\begin{eqnarray}\lim_{x\to+\infty} \frac{f(x)}{x}&=&\lim_{x\to+\infty} \frac{D(x)+mx+q}{x}=\\&=&\lim_{x\to+\infty} \frac{D(x)}{x} +m = m \tag{2}\label{eq2624:2}\end{eqnarray}

Thus, \eqref{eq2624:2} is a *necessary* condition for the existence of an asymptote. This leads to a straightforward procedure:

- Calculate the limit \eqref{eq2624:2}. If this limit does not exist or it is infinite, then the function does not have an asympote. If the limit exists and is finite, then there
*might be*an asymptote with slope \(m\). - If you found a finite \(m\), then you also need to compute the limit \begin{equation}\lim_{x\to+\infty} \left[f(x)-mx\right].\tag{3}\label{eq2624:3}\end{equation}
*If this limit, too, exists and is finite*(call it \(q\)) you can conclude that the function does have an asympote with equation \(y = mx + q\).

As an exercise, test this method with the function

\[f(x) = (x-2) e^{-\frac1x}.\]

You should find the asymptote \(y=x-3\).

Suppose now the function \(f(x)\) is differentiable, at least in a neighborhood of \(+\infty\). We ask ourselves if there is any connection between the potential slant asyptote and the behavior of the derivative at infinity. By De l’Hôpital’s rule, we know that, if

\begin{equation}\lim_{x\to+\infty} f'(x)\tag{4}\label{eq2624:4}\end{equation}

exists (whether finite or infinite), then it is equal to

\[\lim_{x\to+\infty} = \frac{f(x)}{x}.\]

Therefore,

- If limit \eqref{eq2624:4} exists and is infinite, then, by necessity of condition \eqref{eq2624:2} we conclude that the function
*does not*have an asyptote. - If limit \eqref{eq2624:4} exists and is finite, then the function may or may not have an asymptote, depending on the limit \eqref{eq2624:3}.
- If limit \eqref{eq2624:4} does not exist, again
*no conclusion*can be drawn, and we need to test*both*limits \eqref{eq2624:2} and \eqref{eq2624:3}.

Here are some exercises to conclude our discussion.

- Consider \[f(x) = \log x + x.\]Show that the limit \eqref{eq2624:4} (and thus \eqref{eq2624:2}) exists and is finite, and that, however, \(f(x)\)
*does not have*a slant asympote. - Let \[f(x) = \frac{\sin x^2}{x}+x.\] Show that this function
*has*a slant asymptote even though the limit \eqref{eq2624:4} does not exist. The function is represeted in the Figure below, together with its asymptote.

- Modify the function of the previous exercise, so that it has a slant asymptote, but
*unbounded derivative in every neighboorhood of \(+\infty\).* - In the two previous examples the distance \(D(x)\) between the function and the asymptote
*is not monotonic*. Construct then an example of a function that monotonically tends to its asymptote (that is: \(D(x)\) is monotonic) and yet has*unbounded derivative in every neighborhood of \(+\infty\).*