End Behavior and Asymptotes
When sketching a function, it is useful to know what happens far to the left and far to the right. A graph may approach a fixed height, follow an inclined line, or behave differently at its two ends. Limits describe these patterns precisely and help identify the lines the graph approaches.
An asymptote describes end behavior, not necessarily a barrier the graph can never cross. A function may intersect a horizontal or slant asymptote at a finite value of \(x\) and still approach that line as \(x\) tends to infinity.
Each end is considered separately. A function might have a horizontal asymptote as \(x\to\infty\), a different one as \(x\to-\infty\), or a slant asymptote at one end and a horizontal one at the other. Limits at infinity are taken through values of \(x\) in the domain. The relevant end of the domain must extend without bound.
Horizontal Asymptotes
To test for a horizontal asymptote, evaluate the limit of \(f(x)\) at the end in question. If the limit is a finite number \(L\), the horizontal asymptote there is \(y=L\). The limit may differ at the two ends, so both limits should be checked when the domain extends toward both \(+\infty\) and \(-\infty\).
Worked Example: A Horizontal Asymptote at Both Ends
Find the horizontal asymptote of \(f(x)=\frac{3x^2+2x-1}{x^2+1}\) at each end, and determine which side of the line the graph approaches from.
Divide the numerator and denominator by \(x^2\), the highest power in the denominator:
As \(x\to\infty\) and as \(x\to-\infty\), both \(1/x\) and \(1/x^2\) tend to zero. Therefore, at either end,
The horizontal asymptote at both ends is \(y=3\). To check the side of approach, subtract 3:
The denominator is positive for every real \(x\). For \(x>2\), the numerator is positive, so \(f(x)>3\): the graph approaches \(y=3\) from above as \(x\to\infty\). For \(x<2\), the numerator is negative, so \(f(x)<3\); in particular, the graph approaches \(y=3\) from below as \(x\to-\infty\). Checking confirms that the difference tends to zero: its numerator grows at most linearly while its denominator grows quadratically.
For rational functions, comparing degrees can provide a quick way to anticipate a horizontal asymptote, but the limit is the underlying test. When numerator and denominator have the same degree, the limit is the ratio of their leading coefficients; when the numerator has smaller degree, the limit is zero. These shortcuts follow by dividing through by the highest power in the denominator, as in the example.
Slant Asymptotes
A slant asymptote is a nonhorizontal line that the graph approaches. The defining test at a chosen end is that the vertical difference between the function and the line tends to zero:
Polynomial division is a useful way to find such a line for a rational function. If division gives \(f(x)=mx+b+r(x)\), and the remainder term \(r(x)\) tends to zero at the end being considered, then \(y=mx+b\) is a slant asymptote there.
This also explains why division works: it separates the function into a line and a remaining error. The line is an asymptote precisely when that error tends to zero. For a rational function whose numerator’s degree is exactly one greater than its denominator’s degree, polynomial division produces a linear quotient and a proper rational remainder. At an unbounded end, that remainder tends to zero, so the quotient line is a slant asymptote.
Worked Example: A Slant Asymptote from Polynomial Division
Find the slant asymptote of \(f(x)=\frac{2x^2+3x+1}{x-1}\) at both ends, and determine the side of approach.
Divide \(2x^2+3x+1\) by \(x-1\). The first quotient term is \(2x\), since \(2x^2/x=2x\). Multiplying and subtracting gives
The next quotient term is \(5\). Subtracting its product with the divisor leaves
Thus the quotient is \(2x+5\) and the remainder is 6:
Subtract the candidate line. The error is \(6/(x-1)\), which tends to zero at both ends, so the slant asymptote at both \(+\infty\) and \(-\infty\) is \(y=2x+5\). For \(x>1\), \(6/(x-1)>0\), so the graph lies above the line and approaches it from above as \(x\to\infty\). For \(x<1\), \(6/(x-1)<0\), so the graph lies below the line and approaches it from below as \(x\to-\infty\).
Check the division by multiplying the quotient by the divisor and adding the remainder:
This reproduces the original numerator, and the remainder term tends to zero at both ends.
Finding a Line by Limits
When an equation is not easy to divide, limits can test a proposed slant line directly. If a slant asymptote \(y=mx+b\) exists at one end, then \(f(x)=mx+b+\varepsilon(x)\), where \(\varepsilon(x)\to0\). Dividing by \(x\) gives \(f(x)/x=m+b/x+\varepsilon(x)/x\). Since \(b/x\to0\) and \(\varepsilon(x)/x\to0\), the slope can be found from \(\lim f(x)/x=m\), if that finite limit exists. Then test the intercept using \(\lim(f(x)-mx)=b\). These limits must be evaluated at the end being studied; their existence is not guaranteed for every function.
Worked Example: Different Asymptotes at the Two Ends
Find the asymptote at each end of \(f(x)=\frac{x}{\sqrt{x^2+1}}\), and determine the side of approach.
As \(x\to\infty\), factor \(x^2\) inside the square root. Since \(x\) is positive at this end, \(\sqrt{x^2}=x\):
So \(y=1\) is the horizontal asymptote at \(+\infty\). For \(x>0\), \(x^2+1>x^2\), hence \(\sqrt{x^2+1}>x\) and \(f(x)<1\). The graph approaches \(y=1\) from below.
As \(x\to-\infty\), \(\sqrt{x^2}=|x|=-x\), so
So \(y=-1\) is the horizontal asymptote at \(-\infty\). For \(x<0\), \(\sqrt{x^2+1}>|x|=-x\), which implies \(x/\sqrt{x^2+1}>-1\). The graph approaches \(y=-1\) from above. The two limits and the inequalities confirm both asymptotes and both sides of approach.
A Slant Asymptote at Each End
A graph need not have the same asymptote at its two ends. The absolute value implicit in \(\sqrt{x^2}=|x|\) can make a square-root function behave differently on the left and right.
Worked Example: A Square-Root Function with Two Slant Asymptotes
Find the asymptotes of \(g(x)=\sqrt{x^2+4x+5}\) at both ends and determine the side of approach.
Rewrite the expression under the square root as a completed square:
As \(x\to\infty\), the candidate line is \(y=x+2\), because \(x+2\) is positive and grows without bound. Subtract it and rationalize:
The denominator grows without bound as \(x\to\infty\), so the difference tends to zero. It is positive, so \(g(x)\) approaches \(y=x+2\) from above.
As \(x\to-\infty\), the candidate line is \(y=-x-2\), since \(|x+2|=-(x+2)\) for \(x<-2\). At that end,
For \(x<-2\), the denominator on the right is positive and grows without bound as \(x\to-\infty\). Thus the difference tends to zero and remains positive. The graph approaches \(y=-x-2\) from above. In both rationalizations, multiplying the numerator and denominator by the conjugate gives the numerator \(((x+2)^2+1)-(x+2)^2=1\), verifying the displayed equality.
Approach Direction and Common Pitfalls
To decide whether the graph is above or below an asymptote, examine the difference between the function and the asymptote. A positive difference means the function’s \(y\)-value is greater, so the graph is above the line; a negative difference means it is below. The sign must be checked near the end in question, not merely at an arbitrary point in the domain. If the difference changes sign infinitely often while tending to zero, the graph may approach the line without staying on one side.
A related sign error can occur with slant asymptotes. In the division example, the difference from \(y=2x+5\) is \(6/(x-1)\), not \(6/(1-x)\). For \(x<1\), its denominator \(x-1\) is negative, so the difference is negative and the graph is below the line on the left. Checking the subtraction or multiplying the division back prevents this error.
Finally, do not interpret “asymptote” as “a line the graph cannot meet.” The definition concerns a limit at an unbounded end. A graph can cross its asymptote at a finite \(x\)-value and still approach it as \(x\to\infty\) or \(x\to-\infty\). For a careful sketch, record the asymptote and the end behavior separately for each end.
Evaluate limits as \(x\to\infty\) and \(x\to-\infty\) wherever the domain extends toward those ends.
A finite limit of \(f(x)\) gives a horizontal asymptote. To test a slant line, check whether \(f(x)-(mx+b)\to0\), with \(m\ne0\).
Find the sign of the difference \(f(x)-L\) or \(f(x)-(mx+b)\) near the end being studied.
Check Your Understanding
Use limits and differences from the candidate lines to answer these questions.
- If \(\lim_{x\to-\infty}f(x)=4\), what is the horizontal asymptote at the left end?
- For \(f(x)=\frac{x^2+1}{x}\), find the slant asymptote at each unbounded end and verify that the difference tends to zero.
- If \(f(x)-(2x-3)\) is positive for all sufficiently large \(x\) and tends to zero, from which side does the graph approach \(y=2x-3\) as \(x\to\infty\)?
- Can a graph cross a horizontal asymptote at a finite value of \(x\)? Explain using the definition of an asymptote.
- For \(f(x)=\frac{2x^2+3x+1}{x-1}\), is the graph above or below its slant asymptote when \(x<1\)?
