End Behavior Does Not Forbid an Intersection
A horizontal asymptote describes what happens to a function’s values as the input grows without bound in a specified direction. It does not say that the graph must stay on one side of the line, or that it cannot meet the line at a finite input.
For example, a function can approach \(y=2\) as \(x\) tends toward positive and negative infinity, while equaling \(2\) at one or more particular inputs. To analyze such a graph, keep two questions separate: what is the limit at infinity, and where, if anywhere, does the function equal the asymptote?
The definition is about a limit, not about equality at finite inputs. A finite intersection with the line occurs when there is an input \(c\) in the domain such that \(f(c)=L\). If the function’s values are on opposite sides of \(L\) just before and just after \(c\), the graph crosses the line there. A graph can also meet the line without changing sides, so intersection and crossing are not always identical.
Why the Limit Allows a Crossing
Suppose \(f\) has horizontal asymptote \(y=L\) at an end of its domain. Define the difference from the asymptote by \(g(x)=f(x)-L\). The limit condition says that \(g(x)\) tends to \(0\) at that end. It does not require \(g(x)\) to be nonzero for every finite \(x\). In particular, \(g(c)=0\) is entirely possible.
At any finite input \(c\), the graph meets \(y=L\) exactly when
These are different kinds of information: the limit describes values far out toward an end, while the equation \(g(c)=0\) describes a value at a particular input. For a crossing, \(g\) must also take opposite signs on the two sides of the intersection. The limit alone neither guarantees nor rules out that sign change.
Worked Example: A Rational Function Crosses Its Horizontal Asymptote
Worked Example: A Rational Function Crosses Its Horizontal Asymptote
Find the horizontal asymptote of \(f(x)=\frac{2x^2+3x}{x^2+1}\), then determine whether the graph crosses it.
Step 1: Find the end behavior. Divide the numerator and denominator by \(x^2\):
As \(x\to\infty\) or \(x\to-\infty\), \(3/x\to0\) and \(1/x^2\to0\). Therefore, the quotient tends to \(2/1=2\), so \(y=2\) is a horizontal asymptote at both ends.
Step 2: Find possible intersections. Solve \(f(x)=2\), with \(x^2+1\ne0\) for every real \(x\):
Check the function value at that input:
Step 3: Check that it crosses. Subtract the asymptote value:
The denominator is positive for every real \(x\). The numerator is negative when \(x<2/3\) and positive when \(x>2/3\). For a numerical check on either side,
Thus the function meets \(y=2\) at \(x=2/3\) and changes from below the line to above it. The equality check confirms the intersection, and the sign calculation confirms the crossing.
Worked Example: A Trigonometric Function Crosses Repeatedly
Worked Example: A Trigonometric Function Crosses Repeatedly
For \(f(x)=1+\frac{\sin x}{x}\), defined when \(x\ne0\), find the horizontal asymptote as \(x\to\infty\) and show that the graph crosses it at \(x=\pi\).
Step 1: Calculate the limit. Since \(-1\le\sin x\le1\), for \(x>0\) we have
Both bounding expressions tend to \(0\) as \(x\to\infty\). The Squeeze Theorem gives \(\lim_{x\to\infty}\sin x/x=0\), and hence
Therefore \(y=1\) is a horizontal asymptote at the positive end.
Step 2: Check the finite input. Because \(\sin\pi=0\),
The graph meets its horizontal asymptote at \(x=\pi\). To verify that it crosses there, take any \(\delta\) with \(0<\delta<\pi\). Then both \(\pi-\delta\) and \(\pi+\delta\) are positive, so their denominators are positive, while
Consequently, \(f(\pi-\delta)>1\) and \(f(\pi+\delta)<1\). The function takes values on opposite sides of the asymptote on the two sides of \(\pi\), confirming a crossing. The limit at infinity and the finite crossing are compatible: the limit concerns inputs growing without bound, not the value at \(\pi\).
Worked Example: A Horizontal Asymptote the Graph Never Meets
Worked Example: A Horizontal Asymptote the Graph Never Meets
Consider \(h(x)=\frac{x^2-1}{x^2+1}\). Find its horizontal asymptote and determine whether it ever intersects that line.
Step 1: Find the end behavior. Dividing the numerator and denominator by \(x^2\) gives
As \(x\to\infty\) and as \(x\to-\infty\), \(1/x^2\to0\). Thus \(h(x)\to1\), and \(y=1\) is a horizontal asymptote at both ends.
Step 2: Compare the function to the line. Subtract \(1\):
For every real \(x\), \(x^2+1>0\), so \(h(x)-1<0\). Therefore \(h(x)<1\) for every finite input; the graph approaches the asymptote from below but never meets it. For instance, \(h(0)=-1\), and direct comparison confirms \(-1<1\). The function’s end behavior establishes the asymptote, while the strict inequality rules out any finite intersection.
A Reliable Method for Checking a Crossing
When a problem asks whether a function crosses its horizontal asymptote, use the limit and the finite-input equation as separate steps. Finding the asymptote does not by itself answer whether a crossing occurs.
Calculate the limit as \(x\to\infty\), \(x\to-\infty\), or both, as requested. A finite limit \(L\) gives the corresponding horizontal asymptote \(y=L\).
Set \(f(x)=L\), solve for finite inputs, and check that each solution is in the domain.
Compare \(f(x)-L\) on either side of a candidate input. Opposite signs indicate a crossing; a sign that stays the same can indicate a touch without a crossing.
For rational functions, simplifying \(f(x)-L\) often makes the sign test direct. Always keep domain restrictions in view: an algebraic solution to \(f(x)=L\) is an intersection only if the function is actually defined at that input.
Common Mistake: Treating an Asymptote as a Barrier
A frequent incorrect claim is: “If \(y=2\) is a horizontal asymptote, then \(f(x)\) can never equal \(2\).” That conclusion does not follow from the definition. The rational function in the first example is a counterexample: its limit is \(2\), and its value at \(x=2/3\) is also exactly \(2\).
The mistake is to replace a statement about a limit with a statement about every finite input. The definition says that for inputs sufficiently far toward the specified end, the function’s values can be made as close to \(L\) as desired. It does not assert that \(f(x)\ne L\) at all finite inputs. Conversely, a horizontal asymptote does not guarantee a crossing either: \(h(x)=(x^2-1)/(x^2+1)\) approaches \(1\) but remains strictly below it.
Check Your Understanding
Answer each question using the distinction between a limit at infinity and a finite-input equation.
- If \(\lim_{x\to\infty}f(x)=4\), what line is a horizontal asymptote at the positive end?
- For \(p(x)=3+\frac{x+1}{x^2+1}\), find the horizontal asymptote as \(x\to\infty\), then solve \(p(x)=3\).
- For the function in question 2, use the sign of \(p(x)-3\) to decide whether it crosses the asymptote at its finite intersection.
- Can a graph approach a horizontal asymptote without ever meeting it? Give a reason or an example.
- What additional check distinguishes a crossing from an intersection that does not change sides?
