Introduction
A horizontal asymptote describes what a function’s values approach far to the left or far to the right. It does not say that the graph must stay on one side of the asymptote, or that the graph can never meet it at a finite input.
A curve can approach a horizontal line as \(x\) becomes very large in magnitude, cross that line one or more times at finite inputs, and then continue toward the line. To sketch such a curve accurately, keep two kinds of information distinct: the function’s behavior at finite inputs and its limiting behavior at the ends of its domain.
What a Horizontal Asymptote Says
For example, \(\lim_{x\to\infty}f(x)=L\) means that the function values can be made as close to \(L\) as desired by taking \(x\) sufficiently large. This is an end-behavior statement. It does not specify the value of \(f(x)\) at any particular finite input.
Consequently, the equation \(f(c)=L\) for a finite number \(c\) is entirely compatible with \(y=L\) being a horizontal asymptote. Such an equation identifies a point where the graph meets the asymptote. To determine whether the graph crosses there, we need more information, especially continuity and the signs of \(f(x)-L\) nearby.
Finding Finite Intersections with the Asymptote
To find where the graph meets the horizontal line \(y=L\), solve \(f(x)=L\), subject to the domain of \(f\). Equivalently, solve \(f(x)-L=0\). For a rational function, retain the original denominator restrictions: any candidate input where the function is undefined is not an intersection of its graph with the line.
The expression \(f(x)-L\) is particularly useful. When it is positive, the graph lies above \(y=L\); when it is negative, the graph lies below. Its sign can also reveal the side from which the function approaches the asymptote at each end.
Worked Example: An Asymptote the Graph Never Meets
Consider \(f(x)=\dfrac{x^2-1}{x^2+1}\). The denominator is positive for every real \(x\), so the function is defined and continuous everywhere. Dividing numerator and denominator by \(x^2\) gives, for nonzero \(x\),
As \(x\to\infty\) or \(x\to-\infty\), \(1/x^2\to0\), so both limits of \(f(x)\) are \(1\). Thus \(y=1\) is a horizontal asymptote at both ends. To check for finite intersections, subtract \(1\):
Since \(x^2+1>0\), this difference is negative for every real \(x\). It never equals zero. The graph stays below \(y=1\) and never meets the asymptote. This also checks the direction of approach at both ends: it approaches \(1\) from below.
Crossing Requires Careful Language
If \(f(c)=L\), the graph intersects \(y=L\) at \(x=c\), but intersection alone does not establish crossing. The function might stay on the same side of the line near \(c\), or it might be discontinuous there. A sign change in \(f(x)-L\) is evidence about which side the graph occupies, but without continuity it cannot rule out a jump.
Here “sufficiently near” means that there is some positive distance \(\delta\) such that the stated sign condition holds whenever \(0<|x-c|<\delta\), on the specified side or sides. The strictness matters: if there are other intersections arbitrarily close to \(c\), the simple same-side or opposite-side description may not apply.
Why the Continuity Condition Matters
Set \(g(x)=f(x)-L\). The graph is above the line when \(g(x)>0\) and below it when \(g(x)<0\). If \(f\) is continuous near \(c\), then \(g\) is continuous there as well. When \(g(c)=0\) and \(g\) has opposite strict signs immediately to the left and right, the graph is unbroken at the intersection and changes sides there. This is the local reason the sign test works for a continuous function.
Without continuity, a function can change from positive to negative values by jumping over zero, even if it is assigned the value zero at the jump. Conversely, a function can equal zero at one isolated input while remaining positive on both sides. In either situation, sign information does not justify describing the graph as a continuous curve crossing or touching the line.
For a rational function, continuity holds wherever its denominator is nonzero. Thus the crossing test is available at any point in its domain, but it must not be applied at a point where the function is undefined.
Worked Crossings and End Behavior Together
Worked Example: One Crossing and Opposite End Approaches
Consider \(f(x)=3+\dfrac{x}{x^2+1}\). The denominator is positive for all real \(x\), so \(f\) is continuous everywhere. Since \(x/(x^2+1)\to0\) as \(x\to\infty\) and as \(x\to-\infty\), the horizontal asymptote at both ends is \(y=3\).
To find finite intersections with this line, subtract \(3\) and solve:
The denominator is nonzero at \(x=0\), and \(f(0)=3\), so the graph intersects the asymptote at \((0,3)\). Because \(x^2+1>0\), the sign of \(f(x)-3\) is the sign of \(x\): it is negative for \(x<0\) and positive for \(x>0\). Continuity and this opposite-side sign pattern show that the graph crosses \(y=3\) at \(x=0\).
The same expression determines the end approaches. For large positive \(x\), the difference \(x/(x^2+1)\) is positive, so the graph approaches \(3\) from above as \(x\to\infty\). For large negative \(x\), the difference is negative, so it approaches \(3\) from below as \(x\to-\infty\).
Derivative information can help make the sketch consistent with these facts. Differentiating gives
The denominator is positive. The numerator is negative for \(x<-1\), positive for \(-1<x<1\), and negative for \(x>1\). Thus the function decreases, then increases, then decreases, with a local minimum at \(x=-1\) and a local maximum at \(x=1\). Their values are
A consistent sketch therefore has the graph approaching \(y=3\) from below on the left, reaching the local minimum \((-1,5/2)\), crossing the line at \((0,3)\), reaching the local maximum \((1,7/2)\), and then approaching \(y=3\) from above on the right. As a direct check, \(f(0)=3+0/(0^2+1)=3\), as required.
Worked Example: Two Crossings of the Same Asymptote
Consider \(f(x)=4+\dfrac{x^2-1}{(x^2+1)^2}\). The denominator is positive for every real input, so the function is continuous everywhere. The fraction tends to zero as \(x\to\infty\) and as \(x\to-\infty\), so the horizontal asymptote at both ends is \(y=4\).
Subtract the asymptote and solve for intersections:
The denominator does not vanish at either input. Substitution confirms both intersection points:
Since the denominator in \(f(x)-4\) is positive, its sign is the sign of \(x^2-1=(x-1)(x+1)\). This is positive for \(x<-1\), negative for \(-1<x<1\), and positive for \(x>1\). The graph is therefore above the asymptote to the left of \(-1\), below it between the two intersections, and above it to the right of \(1\). Continuity shows that it crosses at both \(x=-1\) and \(x=1\). At both ends the difference tends to zero through positive values, so the graph approaches \(y=4\) from above as \(x\to-\infty\) and as \(x\to\infty\).
For additional sketch information, differentiate the difference term:
The critical inputs are \(x=-\sqrt{3},0,\sqrt{3}\). The derivative is positive for \(x<-\sqrt{3}\), negative for \(-\sqrt{3}<x<0\), positive for \(0<x<\sqrt{3}\), and negative for \(x>\sqrt{3}\). The corresponding function values are
Thus a sketch must include local maxima at \((-\sqrt{3},33/8)\) and \((\sqrt{3},33/8)\), a local minimum at \((0,3)\), the two crossings at \((-1,4)\) and \((1,4)\), and approach to \(y=4\) from above at both ends. These features agree: the graph descends through \(y=4\) at \(-1\), reaches its minimum, rises through \(y=4\) at \(1\), and then turns downward toward the asymptote.
A Common Mistake: Treating an Asymptote as a Barrier
A mistaken argument might say, “The limit is \(L\), so the function cannot equal \(L\) at a finite input.” This confuses a statement about inputs tending to infinity with a statement about every finite input. For instance, in the preceding example, \(\lim_{x\to\infty}f(x)=4\) and \(\lim_{x\to-\infty}f(x)=4\), yet \(f(-1)=f(1)=4\).
Another mistake is to infer crossing from opposite signs without checking continuity. For example, define \(f(x)=1\) for \(x<0\), \(f(0)=0\), and \(f(x)=-1\) for \(x>0\). The signs on the two sides of \(0\) are opposite, but the function is discontinuous there. The graph jumps from one side of \(y=0\) to the other; continuity, which would support the crossing test, is absent. Always check the domain and continuity before interpreting signs as a crossing of an unbroken graph.
Building a Consistent Sketch
When a horizontal asymptote and possible finite crossings are part of a sketch, organize the information rather than letting one feature override another. A reliable sequence is:
Identify the horizontal asymptote at each end where a finite limit exists.
Solve \(f(x)=L\) for each asymptote \(y=L\), excluding inputs outside the domain.
Study the sign of \(f(x)-L\) on the intervals separated by the intersections.
Use the local sign pattern to identify a crossing only when the function is continuous there.
Place known intercepts, extrema, and changes in monotonicity consistently with the crossings and end approaches.
This process matters because an end limit alone cannot tell you whether the graph approaches from above or below, and it cannot reveal finite intersections. The difference \(f(x)-L\) connects these questions: its zeros locate intersections, its sign gives the graph’s side of the line, and its limiting behavior helps describe the approach at infinity.
Check Your Understanding
Use the definitions and sign reasoning in this tutorial to answer the following questions.
- What does it mean for \(y=L\) to be a horizontal asymptote as \(x\to-\infty\)?
- If \(f(x)-L\) is positive for every real \(x\), can the graph intersect \(y=L\) at a finite input? Explain.
- For a continuous function with \(f(c)=L\), what local sign pattern indicates that the graph crosses \(y=L\) at \(c\)?
- Why is continuity needed before interpreting opposite signs on either side as a crossing?
- How can the sign of \(f(x)-L\) tell you whether the graph approaches its horizontal asymptote from above or below?
