Why Test a Limit With Sequences?
A function can behave differently along different routes toward the same input. A graph or a table may suggest a trend, but a carefully chosen list of inputs can make that contrast explicit. The central test is simple: if two sequences of inputs both approach the same target, but the corresponding function values approach different numbers, then the function cannot have one limit at that target.
This is a way to disprove a proposed limit, not a shortcut for proving one. Two sequences that produce the same output trend show only what happens along those two lists of inputs. Other inputs, not sampled by either list, may behave differently.
Input Sequences Approaching a Target
A sequence of inputs is a list \(x_1,x_2,x_3,\ldots\) of numbers. We say that \(x_n\) approaches \(c\), and write \(x_n\to c\), when its terms get arbitrarily close to \(c\) as the index \(n\) increases. For a limit of \(f(x)\) as \(x\to c\), the inputs in the sequence must be in the domain of \(f\) and must not equal \(c\).
The corresponding sequence of function values is \(f(x_1),f(x_2),f(x_3),\ldots\). If these values approach a number \(L\), we write \(f(x_n)\to L\). The key point is that an input sequence approaching \(c\) gives one particular way of examining the function near \(c\).
The two sequences can approach from opposite sides, from the same side, or in any other way allowed by the domain. What matters is that both approach the same target, while their function values have incompatible limiting behavior.
Why Conflicting Sequences Rule Out a Limit
Suppose a function really does approach one number \(L\) as its input approaches \(c\). Then, sufficiently near \(c\), its values must stay as close to \(L\) as we require. In particular, this must happen for inputs taken from any sequence approaching \(c\). Once the sequence inputs are sufficiently close to \(c\), their function values must be close to \(L\) as well.
Thus any valid input sequence approaching \(c\) must produce function values approaching that same \(L\). If one sequence instead approaches \(A\) and another approaches \(B\ne A\), the supposed single limiting value would have to equal both \(A\) and \(B\), which is impossible.
Each must approach the target \(c\), remain in the function’s domain, and avoid \(c\).
Compute \(f(x_n)\) and \(f(y_n)\) using the definition of the function.
If the output sequences approach different numbers, the proposed function limit is impossible.
Worked Examples
Worked Example: Opposite Signs Near Zero
Test whether \(f(x)=\frac{x}{|x|}\) has a limit as \(x\to0\). The function is defined for \(x\ne0\). Use the sequences \(x_n=\frac1n\) and \(y_n=-\frac1n\), where \(n\in\mathbb N\).
First, both input sequences approach zero: as \(n\) increases, \(\frac1n\to0\) and \(-\frac1n\to0\). Their terms are nonzero and are in the domain.
For the first sequence, \(x_n=\frac1n>0\), so \(\left|x_n\right|=\frac1n\). Therefore:
Every term of the first output sequence equals \(1\), so \(f(x_n)\to1\).
For the second sequence, \(y_n=-\frac1n<0\), so \(\left|y_n\right|=\frac1n\). Therefore:
Every term of the second output sequence equals \(-1\), so \(f(y_n)\to-1\). Since \(1\ne-1\), the two sequences have different output limits. Hence \(\lim_{x\to0}\frac{x}{|x|}\) does not exist. As a check, the two computed outputs are exactly the function’s values on positive and negative inputs, respectively; they cannot both approach a single number.
Worked Example: Rational and Irrational Inputs
Define \(f(x)=1\) when \(x\) is rational and \(f(x)=0\) when \(x\) is irrational. Test whether \(f\) has a limit as \(x\to0\). Use \(x_n=\frac1n\) and \(y_n=\frac{\sqrt{2}}{n}\), for \(n\in\mathbb N\).
Both sequences approach zero: \(\frac1n\to0\), and multiplying these terms by the fixed number \(\sqrt{2}\) still gives \(\frac{\sqrt{2}}n\to0\). Neither sequence contains zero.
Each \(x_n=\frac1n\) is rational, so \(f(x_n)=1\) for every \(n\). Thus \(f(x_n)\to1\).
Each \(y_n=\frac{\sqrt{2}}n\) is irrational. Indeed, if \(\frac{\sqrt{2}}n\) were rational, multiplying it by the integer \(n\) would make \(\sqrt{2}\) rational, which it is not. Thus \(f(y_n)=0\) for every \(n\), and \(f(y_n)\to0\).
The input sequences approach the same target, but the output limits are \(1\) and \(0\). Therefore, \(\lim_{x\to0}f(x)\) does not exist. Checking the function’s definition confirms that the outputs are \(1\) along the rational sequence and \(0\) along the irrational sequence.
Worked Example: Two Matching Sequences Do Not Prove a Limit
Consider \(f(x)=\sin(1/x)\), defined for \(x\ne0\), and the two input sequences \(x_n=\frac1{n\pi}\) and \(y_n=\frac1{2n\pi}\), where \(n\in\mathbb N\). Both approach zero, since their denominators grow without bound.
For the first sequence, \(1/x_n=n\pi\). Hence:
so \(f(x_n)\to0\). For the second sequence, \(1/y_n=2n\pi\), and therefore:
so \(f(y_n)\to0\) as well. These calculations agree, but they do not establish that the function limit is zero—or that any limit exists.
To check whether the agreement settles the question, test a third sequence, \(z_n=\frac{1}{\pi/2+2n\pi}\). Its denominator grows without bound, so \(z_n\to0\), and every term is nonzero. Moreover, \(1/z_n=\pi/2+2n\pi\), giving:
Thus \(f(z_n)\to1\), which conflicts with the output limit \(0\) found along the first sequence. The function has no limit at zero. The check is direct: the two original sequences really do give zero, but a third valid sequence gives one. Agreement along two selected sequences was not enough.
What the Test Can and Cannot Establish
A pair of sequences with different output limits is decisive evidence: it rules out a limit, provided the sequences satisfy the conditions of the test. No graphing or additional sampling can rescue the proposed limit, because the function is already shown to approach incompatible values along valid inputs.
The reverse conclusion does not follow. If two, or even many, sequences give the same output trend, that does not show that all inputs near the target behave that way. A sequence test that finds a conflict proves failure; a sequence test that fails to find a conflict is inconclusive. For a positive proof, one needs an argument that accounts for all sufficiently nearby inputs, not just selected lists.
Choosing sequences is therefore a matter of looking for meaningful differences in the function’s behavior. Useful choices include positive and negative inputs, rational and irrational inputs when the function distinguishes them, or values that make part of a formula especially simple. The sequences need not be equally spaced, but they must approach the same target and satisfy the domain restrictions.
Choosing and Checking Sequences
Before drawing a conclusion, verify every part of the test. A sequence that includes the target input is not suitable for a limit test, even if the function is defined there: a limit concerns nearby inputs other than the target itself. Likewise, a sequence that includes inputs outside the domain cannot be used to evaluate the function. Finally, the input sequences must approach the same target; comparing outputs from sequences heading toward different inputs answers a different question.
When the conditions hold, calculate the output values rather than relying on an impression from the input formulas. The function may simplify in different ways on different sequences, as in the sign example, or assign different values based on a property such as rationality. State the limit of each output sequence explicitly, then compare the resulting numbers.
This method turns the idea of a limit into a precise test of consistency. A genuine function limit must be compatible with every way of approaching its input. Two carefully chosen sequences can reveal when that consistency fails.
Check Your Understanding
For each question, focus on whether the sequences meet the test’s conditions and what their output behavior can establish.
- What conditions must two input sequences satisfy to test a proposed limit at \(c\)?
- If the function values along the two sequences approach different numbers, what can you conclude about the function limit?
- Why does finding the same output limit along two sequences not prove that the function has a limit?
- For \(f(x)=x/|x|\), what output limits result from \(x_n=1/n\) and \(y_n=-1/n\)?
- Can a sequence containing the target input be used in this limit test? Explain why or why not.
