Why One-Sided Behavior Matters
A graph can approach a vertical line while its function values grow without bound. At an excluded input, the behavior on the left may differ from the behavior on the right: one side may rise without bound while the other falls, or both sides may head in the same direction. To describe the graph accurately, we must examine each side separately.
The domain tells us where the function is defined. An excluded value divides the domain into intervals, and the graph on each interval forms a separate branch. One-sided limits describe how those branches behave near the excluded input. Derivative signs can then tell us about concavity on each branch.
Definition of a Vertical Asymptote
Let \(f\) be defined on at least one side of \(x=a\), possibly with \(f(a)\) undefined. The vertical line \(x=a\) is a vertical asymptote if at least one of the following one-sided limits is \(+\infty\) or \(-\infty\):
A one-sided limit considers inputs approaching \(a\) from only one direction. The notation \(x\to a^-\) means \(x<a\) and \(x\) approaches \(a\); \(x\to a^+\) means \(x>a\) and \(x\) approaches \(a\). If the domain includes only one side of \(a\), only that side needs to be considered.
The limit \(+\infty\) means that \(f(x)\) becomes greater than any chosen positive bound as \(x\) approaches \(a\) from the specified side. The limit \(-\infty\) means that \(f(x)\) becomes less than any chosen negative bound. These are descriptions of unbounded behavior, not finite values of the function.
Why an Infinite One-Sided Limit Gives an Asymptote
Suppose, for example, that \(\lim_{x\to a^+} f(x)=+\infty\). By the definition of this limit, for every positive number \(M\), there is a distance \(\delta>0\) such that \(a<x<a+\delta\) implies \(f(x)>M\). At the same time, those inputs approach \(a\), so their horizontal distance from the line \(x=a\) approaches zero.
Thus, along this branch, the graph gets arbitrarily close horizontally to \(x=a\), while its height grows beyond every bound. That is precisely the behavior described by a vertical asymptote. The same reasoning applies to a limit of \(-\infty\), and to a left-sided limit. A complete check therefore asks separately what happens as \(x\) approaches \(a\) from each side available in the domain.
Find an excluded input \(a\) and identify which intervals of the domain approach it.
Determine whether \(f(x)\) tends to a finite value, \(+\infty\), or \(-\infty\) on each available side.
Record the direction of unbounded behavior on each side. Do not assume the two sides behave alike.
Use the sign of \(f''\) on each domain interval; the excluded input itself is not part of either interval.
Worked Example: Opposite Directions on the Two Sides
Worked Example: Opposite Directions on the Two Sides
Analyze the vertical asymptote and concavity of \(f(x)=\frac{1}{x-2}\).
Step 1: Find the domain intervals. The denominator is zero at \(x=2\), so \(f\) is undefined there. Its domain is \((-\infty,2)\cup(2,\infty)\). The two branches lie on opposite sides of \(x=2\).
Step 2: Compute the one-sided limits. As \(x\to2^-\), \(x-2\) is negative and approaches zero. Therefore its reciprocal decreases without bound:
As \(x\to2^+\), \(x-2\) is positive and approaches zero, so its reciprocal increases without bound:
At least one one-sided limit is infinite, so \(x=2\) is a vertical asymptote. The left branch falls without bound; the right branch rises without bound.
Step 3: Determine concavity on both intervals. Rewrite \(f(x)=(x-2)^{-1}\). Differentiating gives
For \(x<2\), the denominator of \(f''\) is negative, so \(f''(x)<0\); the branch is concave down on \((-\infty,2)\). For \(x>2\), the denominator is positive, so \(f''(x)>0\); the branch is concave up on \((2,\infty)\).
Check: Differentiating \(-\frac{1}{(x-2)^2}\) gives \(-[-2(x-2)^{-3}]=2(x-2)^{-3}\), confirming the second derivative and its signs. The one-sided limits also agree with the signs of the denominator: negative immediately to the left of \(2\), positive immediately to the right.
Concavity Belongs to Domain Intervals
The sign of \(f''\) describes concavity on an interval where the function is defined and twice differentiable. If an excluded input separates the domain, make a separate concavity decision on each interval. A sign chart is useful because the sign of \(f''\) may be the same on both sides or may change.
In particular, a vertical asymptote does not force the two branches to have opposite concavity. Nor does a change in concavity by itself establish a vertical asymptote: the asymptote question is answered by one-sided limits, while concavity is answered by the second derivative on the intervals in the domain.
Worked Example: Both Sides Rise Without Bound
Worked Example: Both Sides Rise Without Bound
Analyze the vertical asymptote and concavity of \(g(x)=\frac{x+1}{(x-3)^2}\).
Step 1: Identify the domain and limits. The denominator is zero at \(x=3\), and the function is defined for all other real \(x\). As \(x\to3\), the numerator approaches \(4\), while \((x-3)^2\) approaches zero through positive values on both sides. Hence
Therefore \(x=3\) is a vertical asymptote, and both branches rise without bound near it.
Step 2: Find the second derivative. Using the quotient rule,
Differentiate again, writing \(g'(x)=(-x-5)(x-3)^{-3}\):
Step 3: Determine the concavity intervals. The denominator \((x-3)^4\) is positive wherever \(g''\) is defined. Thus the sign of \(g''\) is the sign of \(x+9\). It is negative for \(x<-9\) and positive for \(x>-9\), apart from the excluded value \(x=3\). Consequently \(g\) is concave down on \((-\infty,-9)\), concave up on \((-9,3)\), and concave up on \((3,\infty)\).
Check: The product-rule calculation gives the displayed \(g''\), and its denominator is positive on both sides of \(3\). Near \(3\), \(x+9>0\), so \(g''>0\) on both adjacent intervals, confirming that both branches are concave up near the asymptote even though they are separated there.
An Excluded Input Does Not Always Give an Asymptote
A zero denominator or another domain restriction signals that a value is excluded, but the resulting limit might be finite. If so, there is no vertical asymptote at that input. This distinction is important when simplifying an expression: cancellation can reveal a finite limit, but it does not put the excluded input back into the original function's domain.
Worked Example: A Finite Limit at an Excluded Input
Consider \(h(x)=\frac{x^2-1}{x-1}\), whose domain excludes \(x=1\). Decide whether \(x=1\) is a vertical asymptote.
Factor the numerator for \(x\ne1\):
As \(x\) approaches \(1\) from either side, \(x+1\) approaches \(2\). Therefore
Neither limit is infinite, so \(x=1\) is not a vertical asymptote. The graph follows the line \(y=x+1\) except that the original function has no point at \((1,2)\).
Check: For every \(x\ne1\), the simplified expression has the same value as the original: \((x-1)(x+1)/(x-1)=x+1\). Substituting \(x=1\) into the simplified expression gives \(2\), matching both limits, but this does not change the original function's domain.
When the Domain Has Only One Side
An excluded value can be a boundary of the domain rather than a point with domain values on both sides. In that case, evaluate the one-sided limit from the side that exists. For example, \(q(x)=\ln(x-1)\) is defined only for \(x>1\). As \(x\to1^+\), \(x-1\to0^+\), so \(\ln(x-1)\to-\infty\). Thus \(x=1\) is a vertical asymptote even though there is no left branch in the domain.
Its concavity can also be determined on the available interval. Differentiation gives
So the branch on \((1,\infty)\) is concave down. This conclusion concerns the domain interval that exists; no behavior to the left of \(1\) should be inferred.
Common Pitfall: Treating Every Exclusion as an Asymptote
A common incorrect argument is: “The denominator is zero at \(x=1\), so \(x=1\) must be a vertical asymptote.” For \(h(x)=\frac{x^2-1}{x-1}\), that calculation stops too early. After factoring and canceling for \(x\ne1\), the expression is \(x+1\), and both one-sided limits equal \(2\), not infinity. The correction is to use the domain restriction to identify a candidate, then compute the one-sided limits before deciding.
A second useful check is to keep the two sides separate. A two-sided limit might fail to exist because the left and right limits have different signs of infinity. That does not prevent \(x=a\) from being a vertical asymptote: either one-sided infinite limit is enough.
Check Your Understanding
For each question, distinguish the domain, one-sided behavior, and concavity conclusions where requested.
- For \(f(x)=\frac{1}{x+4}\), find both one-sided limits at \(x=-4\) and state which way each branch goes.
- If \(\lim_{x\to a^-}f(x)=-\infty\) and \(\lim_{x\to a^+}f(x)=+\infty\), is \(x=a\) a vertical asymptote? Explain using the definition.
- Why does an excluded input with two finite one-sided limits fail to be a vertical asymptote?
- For \(p(x)=\ln(x+2)\), which one-sided limit is relevant at the domain boundary \(x=-2\), and what does it imply?
- When an excluded input splits a domain into two intervals, what information does the sign of \(f''\) provide on each interval?
