Why Test One Side at a Time?
A function may grow without bound as \(x\) approaches a particular value. When this happens, the graph can draw closer and closer to a vertical line while rising or falling indefinitely. That line is a vertical asymptote.
The behavior on the two sides of the line need not match. A function might approach \(+\infty\) from one side and \(-\infty\) from the other, or it might become unbounded on only one side. Therefore, deciding whether \(x=a\) is a vertical asymptote requires checking the one-sided limits separately.
In this definition, \(x\to a^-\) means that \(x\) approaches \(a\) through values less than \(a\), and \(x\to a^+\) means that \(x\) approaches through values greater than \(a\). The symbol \(+\infty\) describes values that increase without bound; \(-\infty\) describes values that decrease without bound. Neither is a real number that the function attains as a limit.
What an Infinite One-Sided Limit Means
For example, \(\lim_{x\to a^+}f(x)=+\infty\) means that no matter how large a positive number \(M\) is chosen, \(f(x)\) becomes greater than \(M\) whenever \(x\) is sufficiently close to \(a\) from the right. More precisely, for every \(M>0\), there is a \(\delta>0\) such that
The other three one-sided possibilities are defined by reversing the side condition, the inequality, or both. For instance, a left-hand limit of \(-\infty\) requires that for every \(M>0\), \(f(x)<-M\) whenever \(0<a-x<\delta\).
This definition explains why testing a few values can be suggestive but is not enough to establish an infinite limit. Values that grow in a table might eventually turn around. A limit claim concerns every sufficiently close input, not just a handful of selected inputs.
Worked Example: Both Sides Rise Without Bound
Worked Example: Both Sides Rise Without Bound
Determine whether \(x=1\) is a vertical asymptote of \(f(x)=\dfrac{x+2}{(x-1)^2}\).
First examine the numerator and denominator near \(x=1\). If \(|x-1|<1\), then \(0<x<2\), so \(x+2>2\). Also, for \(x\ne1\), the squared denominator is positive. Thus \(f(x)>0\) on both sides near \(1\).
To verify that the values grow without bound, let \(M>0\) and choose \(\delta=\min\left(1,\sqrt{\dfrac{2}{M}}\right)\). Whenever \(0<|x-1|<\delta\), the numerator is greater than \(2\), while \(0<(x-1)^2<\delta^2\). Therefore,
The last inequality follows because \(\delta\le\sqrt{2/M}\), so \(\delta^2\le 2/M\). The estimate holds for \(x<1\) and for \(x>1\). Hence both one-sided limits are \(+\infty\), and \(x=1\) is a vertical asymptote.
As a numerical check, \(f(0.9)=2.9/0.01=290\), and \(f(1.1)=3.1/0.01=310\). These values agree with the conclusion, although the definition—not the two sample values—establishes it.
Worked Example: A Vertical Asymptote with Opposite Signs
Worked Example: Opposite One-Sided Limits
Determine whether \(x=2\) is a vertical asymptote of \(g(x)=\dfrac{x+1}{x-2}\), and find the direction of each one-sided limit.
Near \(2\), the numerator \(x+1\) is positive. The denominator \(x-2\) is negative to the left of \(2\) and positive to the right. Thus the quotient is negative on the left and positive on the right.
For a magnitude check, suppose \(0<|x-2|<1\). Then \(1<x<3\), so \(x+1>2\). On the right, \(0<x-2<\delta\) implies
Given any \(M>0\), choose \(\delta=\min(1,2/M)\). Then the right-hand values exceed \(M\), so \(\lim_{x\to2^+}g(x)=+\infty\).
On the left, \(0<2-x<\delta\) and \(x+1>2\), giving
With the same choice of \(\delta\), the values are less than \(-M\). Therefore \(\lim_{x\to2^-}g(x)=-\infty\). Since at least one one-sided limit is infinite, \(x=2\) is a vertical asymptote, even though the two sides have opposite signs.
For a numerical check, \(g(1.9)=2.9/(-0.1)=-29\), while \(g(2.1)=3.1/0.1=31\). The signs and sizes agree with the one-sided conclusions.
A Two-Sided Limit Is Not Required
A finite two-sided limit requires the left- and right-hand behavior to approach the same finite value. An infinite two-sided limit likewise requires agreement on the direction of unbounded behavior. But the vertical-asymptote test in this tutorial is deliberately one-sided: either side alone can establish the asymptote.
The reason is geometric. A vertical asymptote describes the graph’s behavior as its input approaches the line’s \(x\)-coordinate. If the graph becomes unbounded as it approaches from even one side, the graph has an unbounded branch near that line. The branch on the other side may behave differently without removing the asymptote.
Worked Example: A Hole Is Not a Vertical Asymptote
Worked Example: Finite Limit at an Undefined Point
Consider \(h(x)=\dfrac{x^2-1}{x-1}\), defined for \(x\ne1\). Decide whether \(x=1\) is a vertical asymptote.
For \(x\ne1\), factor the numerator and simplify:
The cancellation is valid only for \(x\ne1\), which is precisely the domain of the original expression. Thus, for inputs near \(1\) but not equal to \(1\), the function values are given by \(x+1\). It follows that
For example, \(h(0.9)=1.9\) and \(h(1.1)=2.1\), as substitution into the simplified expression confirms. The values stay near \(2\), rather than growing without bound. So \(x=1\) is not a vertical asymptote. The original formula is undefined there, and its graph has a hole at \((1,2)\).
This example highlights an important distinction: being undefined at \(x=a\) does not by itself make \(x=a\) a vertical asymptote. What matters is the limit behavior around \(a\). A missing function value can occur while both one-sided limits remain finite.
Other Finite Discontinuities
A function can also fail to join smoothly at a point while its nearby values remain bounded. For instance, define
For \(x<0\), the function is constantly \(1\), so \(\lim_{x\to0^-}q(x)=1\). For \(x>0\), it is constantly \(2\), so \(\lim_{x\to0^+}q(x)=2\). The two sides approach different finite values; neither tends to \(+\infty\) or \(-\infty\). The function has a finite jump at \(0\), but \(x=0\) is not a vertical asymptote.
Thus, a hole and a jump can both involve a discontinuity, but neither automatically implies an asymptote. For this decision, ask specifically whether at least one one-sided limit is infinite.
A Reliable Test
Look for a point where the function may be undefined or where its formula could become unbounded.
Determine whether \(f(x)\) tends to \(+\infty\), \(-\infty\), or a finite value as \(x\to a^-\).
Make the same determination as \(x\to a^+\). Do not assume it matches the left-hand behavior.
If either side has an infinite limit, \(x=a\) is a vertical asymptote. If both sides have finite limits, it is not.
For rational functions, factoring and checking signs near a denominator zero can help determine the one-sided behavior. The key is to evaluate what the function does for nearby inputs, not merely to note that the denominator is zero at \(a\). A denominator zero may accompany unbounded behavior, but cancellation can instead leave a finite limit, as the hole example showed.
Common Mistake: Treating Every Denominator Zero as an Asymptote
Worked Example: Correcting a Denominator-Zero Test
A tempting but incorrect claim is: “The denominator of \(\dfrac{x^2-1}{x-1}\) is zero at \(x=1\), so \(x=1\) is a vertical asymptote.” The first observation is correct: substituting \(x=1\) into the denominator gives \(1-1=0\). But it does not settle the limit.
For \(x\ne1\), factor and cancel the common factor:
Now take the limit of the expression valid on the punctured neighborhood:
Substituting nearby values checks the result: at \(x=0.99\), the original quotient is \((0.99^2-1)/(0.99-1)=1.99\), and at \(x=1.01\), it is \((1.01^2-1)/(1.01-1)=2.01\). Both are finite and close to \(2\). The corrected conclusion is that \(x=1\) is a hole, not a vertical asymptote.
The cancellation in this calculation does not redefine the original function at \(x=1\). It gives an equivalent expression only for \(x\ne1\), which is enough to compute the limit. The asymptote decision depends on that nearby behavior.
Summary
To determine whether \(x=a\) is a vertical asymptote, test the one-sided limits separately. One infinite one-sided limit is sufficient, and the two sides can diverge in opposite directions. If nearby values instead approach finite numbers, then an undefined point or a finite jump is not a vertical asymptote.
Check Your Understanding
For each question, focus on the behavior from each side of the candidate value.
- If \(\lim_{x\to a^-}f(x)=+\infty\) and \(\lim_{x\to a^+}f(x)=-\infty\), is \(x=a\) a vertical asymptote? Explain.
- Can a function have a hole at \(x=a\) without a vertical asymptote there? What limit behavior would show this?
- For \(r(x)=\dfrac{1}{(x+3)^2}\), determine the signs and infinite-limit behavior on each side of \(x=-3\).
- A function has left- and right-hand limits \(4\) and \(7\) at \(x=0\). Is \(x=0\) a vertical asymptote? Why or why not?
- Why does a zero in the denominator of a rational expression not, by itself, establish a vertical asymptote?
