# Other Asymptotes

A few days ago, a reader asked if you could find the location of horizontal asymptotes from the derivative of a function. The answer, alas, is no. You can determine that a function has a horizontal asymptote from its derivative, but not where it is. This is because a function and its many possible vertical translations have the same derivative but different horizontal asymptotes. To locate it precisely, we need more information, an initial condition (a point on the curve).

I have written a post about the relationship between vertical asymptotes and the derivative of the function here. Today’s post will discuss horizontal asymptotes and their derivatives.

Definition: a straight line associated with a curve such that as a point moves along an infinite branch of the curve the distance from the point to the line approaches zero and the slope of the curve at the point approaches the slope of the line. (Merriam-Webster dictionary). Thus, asymptotes can be vertical, horizontal, or slanted.

Horizontal Asymptotes

Horizontal asymptotes are a form of end behavior: they appear as $\displaystyle x\to \infty$ or $\displaystyle x\to -\infty$. Specifically, if a function has a horizontal asymptote(s), then $\displaystyle \underset{{x\to \pm \infty }}{\mathop{{\lim }}}\,{f}'\left( x \right)=0$,  At an asymptote, the curve must approach a horizontal line, therefore its slope must be approaching zero as $\displaystyle x\to \infty$ or $\displaystyle x\to -\infty$.  The converse is not true: If the derivative approaches zero, the function may not have a horizontal asymptote. A counterexample is $\displaystyle f\left( x \right)=\sqrt[3]{x}$.

In the first four examples that follow, the derivative approaches 0 as $\displaystyle x\to \infty$ and/or $\displaystyle x\to -\infty$

Example 1: $\displaystyle {f}'\left( x \right)=\frac{1}{{1+{{x}^{2}}}}$  (Figure 1 in blue).This is the derivative of $\displaystyle f\left( x \right)={{\tan }^{{-1}}}(x),\ -\tfrac{\pi }{2}. (Figure 1 in red). We see that the derivative approaches zero in both directions, this tells us that there may be horizontal asymptote(s). We must look at the function to find where they are: $\displaystyle y=\tfrac{\pi }{2}$ and $\displaystyle y=-\tfrac{\pi }{2}$

Figure 1

Example 2: 2008 AB 19 $\displaystyle p\left( x \right)=\frac{{5+{{2}^{x}}}}{{1={{2}^{x}}}}$. (Figure 2 in Red)

So many students missed this (pre-calculus ?) question on the exam that the question was not counted! As $\displaystyle x\to \infty$ the 2x terms approach infinity at the same rate so the limit is the ratio of their coefficients: the asymptote is y = –1. And as $\displaystyle x\to -\infty$ the 2x terms approach zero so the fractions approaches an asymptote at y = 5. (There is a vertical asymptote at x = 0, not shown.)

The derivative is $\displaystyle {p}'\left( x \right)=\frac{{6\ln (2)\cdot {{2}^{x}}}}{{{{{\left( {1-{{2}^{x}}} \right)}}^{2}}}}$ (Figure 2 blue). The derivative approaching zero in both directions, but the location of the asymptotes must be determined from the equation of the function.

Figure 2

Example 3: $\displaystyle g\left( x \right)=\frac{{-3}}{{{{{\left( {x+1} \right)}}^{2}}}}$ (Figure 3 in blue). This is the derivative of $\displaystyle g(x)=\frac{3}{{x+1}}+2$ (Figure 2 in red). Notice that replacing the “+2” with a different constant will translate the curve and its asymptote up or down. Here the function has the same asymptote in both directions. (There is a vertical asymptote at x = –1, not shown)

Figure 3

Example 4: Consider the function $\displaystyle s\left( x \right)={{1.1}^{=x}}\sin \left( x \right)$See figure 4.

As $\displaystyle x\to \infty$ this function’s values get closer to zero and the graph appears to have the x-axis as a horizontal asymptote. The interesting thing is that the function’s value at every integral  multiple of π is zero and the function changes sign there. In other words it crosses and re-crosses its asymptote and moves away from and back towards the asymptote. Each time it crosses the asymptote it moves less far from the asymptote, but continues to move towards and away from it. It’s derivative (not shown) exhibits the same phenomenon.

This feature is difficult to see on a graph since the values become exceedingly small in absolute value very quickly.

Figure 4

Slant Asymptotes

Other lines lines may also be asymptotes, called slant asymptotes. If as $\displaystyle x\to \infty$ or $\displaystyle x\to -\infty$, the function approaches a non-horizontal line, which appears as a slant asymptote. In this case the derivative approaches a constant equal to the slope of the line.

Example 5: $\displaystyle {h}'\left( x \right)=1-\frac{1}{{{{{\left( {x-2} \right)}}^{2}}}}$ (Figure 5 in blue) is the derivative of $\displaystyle h\left( x \right)=\frac{{{{x}^{2}}+x-5}}{{(x-2)}}=x+3+\frac{1}{{\left( {x-2} \right)}}$.

The derivative shows a horizontal asymptote at y = 1, so the function will have a slant asymptote with a slope of 1. This is obvious (I hope) from the second form of the function where the final term approaches zero as $\displaystyle x\to \infty$ and $\displaystyle x\to -\infty$. For positive numbers, the last term is positive and the function is therefore above line; for negative numbers the last term is negative and the function is below the line. Vertical translations will move the function and its slant asymptote, but not change the asymptote’s slope. (There is a vertical asymptote at x = 2, not shown)

Figure 5

Other “Asymptotes”

Example 6: Playing with the idea from examples 5, if the function were $\displaystyle k\left( x \right)={{x}^{2}}+x+\frac{1}{{\left( {x-2} \right)}}$. (Red in figure 6). The curve will approach the parabola $\displaystyle y={{x}^{2}}+x$ (Figure 6 black dotted) as an “asymptote.” Its derivative $\displaystyle {k}'\left( x \right)=2x+1-\frac{1}{{{{{\left( {x-2} \right)}}^{2}}}}$ (Figure 6 blue) approaches the line (its own slant asymptote) $\displaystyle y=2x+1$ (not shown). There is a vertical asymptote at x = 2 (not shown).

Figure  6

Another interesting feature of this kind of function is this:  $\displaystyle k\left( x \right)={{x}^{2}}+x+a+\frac{1}{{x-2}}$ is the same function translated up or down by an amount a. The resulting functions have the same derivative and the same parabola for their asymptote! Try investigating this situation using Desmos or GeoGebra or some other graphing utility with a slider.

This type of “asymptote” might make a good topic for an exploration, project, or investigation by a student.

Revised March 23, 2021.