What a Horizontal Asymptote Describes
A graph may approach a fixed height as its input moves farther and farther to the right or to the left. That fixed height is described by a horizontal asymptote. The definition uses a limit at infinity: it describes what happens along an end of the graph, not necessarily what happens at every finite input.
The two directions must be considered separately. A function can approach one height as \(x\) increases without bound and a different height as \(x\) decreases without bound. It can also have a horizontal asymptote in just one direction, or in neither direction.
In precise terms, \(\lim_{x\to+\infty}f(x)=L\) means that for every \(\varepsilon>0\), there is a number \(M\) such that whenever \(x>M\) is in the domain of \(f\), \(|f(x)-L|<\varepsilon\). For \(x\to-\infty\), the condition is that \(x<-M\) implies \(|f(x)-L|<\varepsilon\). In either case, the graph eventually lies as close to \(y=L\) as desired on the specified end.
Finding the Asymptote at Each End
Earlier material on limits at positive and negative infinity established how to calculate these limits. For rational functions, degree comparisons and leading coefficients can often determine a finite limit. Once the limit is known, identifying the corresponding horizontal asymptote is direct: the limit value is the height \(L\) in the line \(y=L\). Remember to calculate the two directional limits separately when the function or its domain permits them to differ.
Worked Example: A Rational Function with the Same Asymptote at Both Ends
Consider \(f(x)=\frac{x}{x^2+1}\). Determine its horizontal asymptote as \(x\to+\infty\) and as \(x\to-\infty\).
The denominator has degree \(2\), while the numerator has degree \(1\). Dividing numerator and denominator by \(x^2\), for \(x\ne0\), gives
As \(x\to+\infty\), \(1/x\to0\) and \(1/x^2\to0\), so the numerator tends to \(0\) and the denominator tends to \(1\). Therefore \(\lim_{x\to+\infty}f(x)=0/1=0\).
As \(x\to-\infty\), \(1/x\to0\) and \(1/x^2\to0\) as well. Thus \(\lim_{x\to-\infty}f(x)=0/1=0\). The graph has the horizontal asymptote \(y=0\) at both ends.
As a check on the function’s values, \(f(1)=1/(1+1)=1/2\), and \(f(-1)=-1/(1+1)=-1/2\). These finite values do not change either limit: the limits describe the distant ends of the graph.
Worked Example: Different Asymptotes at the Two Ends
Define \(g\) by
For \(x\to-\infty\), the applicable formula is \(g(x)=1+1/x\). Since \(1/x\to0\),
For \(x\to+\infty\), the applicable formula is \(g(x)=2+1/x\). Therefore,
The horizontal asymptote on the left is \(y=1\); the horizontal asymptote on the right is \(y=2\). The middle part of the function does not affect either end limit. Checking the relevant branches confirms the calculations: at \(x=-2\), \(g(-2)=1-1/2=1/2\), while at \(x=2\), \(g(2)=2+1/2=5/2\). These values are consistent with the respective tail formulas but are not themselves the limits.
Why Crossing an Asymptote Is Possible
A limit at infinity constrains values sufficiently far out along one end of a graph. It does not assert that \(f(x)\) is always above or always below \(L\), and it says nothing by itself about a particular finite input. Consequently, a graph can meet or cross its horizontal asymptote at a finite point without contradicting the definition.
The construction below makes this explicit. Choose any real number \(L\) and any finite input \(a\), and define \(h(x)=L+\frac{x-a}{x^2+1}\). The denominator \(x^2+1\) is positive for every real \(x\). At \(x=a\), the added fraction is zero, so \(h(a)=L\): the graph meets the line \(y=L\) there. The fraction is negative for \(x<a\) and positive for \(x>a\), so the graph passes from one side of \(y=L\) to the other at that point.
To confirm that \(y=L\) is nevertheless a horizontal asymptote at both ends, consider the added term. For \(|x|\geq1\), the denominator satisfies \(x^2+1\geq x^2\), and \(|x-a|\leq |x|+|a|\). Hence
The right-hand side tends to \(0\) as \(|x|\to\infty\), so the added term tends to \(0\) in both directions. Thus \(h(x)\to L\) as \(x\to+\infty\) and as \(x\to-\infty\), despite the crossing at \(x=a\).
Worked Example: A Rational Graph Crosses Its Horizontal Asymptote
Consider \(r(x)=\frac{2x^2+x}{x^2+1}\). First determine its horizontal asymptotes. Dividing numerator and denominator by \(x^2\) gives, for \(x\ne0\),
As \(x\to+\infty\), the numerator tends to \(2\) and the denominator tends to \(1\), so the limit is \(2/1=2\). As \(x\to-\infty\), the numerator again tends to \(2\) and the denominator again tends to \(1\), so this limit is also \(2\). The horizontal asymptote at both ends is \(y=2\).
To find where the graph meets this line, solve \(r(x)=2\). Because \(x^2+1>0\), multiplication by the denominator is valid:
Substitution checks the result: \(r(2)=(2\cdot4+2)/(4+1)=10/5=2\). To verify that the graph crosses, rather than merely touches, compare its value with \(2\):
The denominator is always positive. Thus \(r(x)-2<0\) when \(x<2\), and \(r(x)-2>0\) when \(x>2\). The graph is below \(y=2\) immediately to the left of \(x=2\) and above it immediately to the right. It crosses its horizontal asymptote at \((2,2)\).
A Common Mistake
A common error is to treat “horizontal asymptote” as if it meant “a line the graph never intersects.” For the function in the worked example, one might reason incorrectly: “The limit is \(2\), so \(r(x)\) cannot equal \(2\).” That conclusion does not follow. In fact, direct substitution gives
The correction is to keep the roles of the statements separate. The limit says that \(r(x)\) gets arbitrarily close to \(2\) as \(x\) tends to either infinity. The equation \(r(2)=2\) describes one finite input. Both statements are true at the same time.
When checking a possible crossing, solve \(f(x)=L\) and verify the solution by substitution. If the question is whether the graph passes from one side of the line to the other, check the sign of \(f(x)-L\) on either side of the intersection. The existence of a horizontal asymptote alone does not settle that sign question.
Why the Direction Matters
Writing down the direction is part of identifying the asymptote correctly. If the limit as \(x\to+\infty\) is \(L\), then \(y=L\) describes the right-hand end. If the limit as \(x\to-\infty\) is \(K\), then \(y=K\) describes the left-hand end. It is possible that \(L=K\), as in the rational examples above, but that need not be true, as the piecewise example shows.
A finite limit at one end does not guarantee a finite limit at the other. Nor does a horizontal asymptote determine whether the graph crosses the line, or how many times it might meet it. These questions require information about the function’s values at finite inputs, not just its end limits.
Check Your Understanding
Use the definition and the distinction between end behavior and finite values to answer these questions.
- If \(\lim_{x\to+\infty}f(x)=5\), what is the horizontal asymptote at the right-hand end?
- Can a function have different horizontal asymptotes as \(x\to-\infty\) and \(x\to+\infty\)? Explain briefly.
- For \(f(x)=\frac{x}{x^2+1}\), identify the horizontal asymptotes and determine whether the graph meets \(y=0\) at a finite input.
- Why does a finite limit at infinity not imply that a function never equals its limit value?
- What additional sign check can show that a graph crosses, rather than merely meets, a horizontal asymptote?
