Why Use Sequences to Test a Derivative?
A derivative at a point is defined by a limit of difference quotients as the increment approaches zero. That limit must give the same finite value no matter how the admissible increments approach zero. This makes sequences a direct way to test whether a derivative exists: if two sequences approach zero but their difference quotients approach incompatible values, no single derivative value can work.
The method is especially useful when a function is defined piecewise, behaves differently on different parts of its domain, or has an oscillatory difference quotient. The key is to choose increments that make the quotient easy to evaluate, and then verify both that the increments are admissible and that they approach zero.
Difference Quotients and Admissible Increments
Let \(f:A\to\mathbb{R}\), where \(A\subseteq\mathbb{R}\), and let \(a\in A\). An increment \(h\) is admissible at \(a\) when \(a+h\in A\). For a nonzero admissible increment, define the difference quotient by
To define a derivative by a limit, there must be domain points other than \(a\) arbitrarily close to \(a\). In other words, \(a\) must be an accumulation point of \(A\setminus\{a\}\). If \(a\) is not such an accumulation point, there are no admissible nonzero increments arbitrarily close to zero. A limit condition would then fail to determine a unique derivative.
The restriction \(a+h\in A\) matters: the function must be defined at the input used in the quotient. When \(A=\mathbb{R}\), every nonzero \(h\) is admissible. If the domain is restricted, only increments that keep \(a+h\) in the domain can be used.
The Sequential Criterion
A limit of \(Q(h)\) exists with value \(L\) precisely when every sequence of admissible nonzero increments tending to zero makes the quotient tend to \(L\). This criterion gives a practical way to disprove a limit: it is enough to find two such sequences whose quotients have different limiting behavior.
Here, the limit on the left is the ordinary finite limit. The theorem is an equivalence: if the limit exists, every allowed sequence must produce it; conversely, if every such sequence produces \(L\), then the limit exists.
For every \(\varepsilon>0\), there is a \(\delta>0\) such that every admissible \(h\) with \(0<|h|<\delta\) satisfies \(|Q(h)-L|<\varepsilon\).
If \(h_n\to 0\), then eventually \(0<|h_n|<\delta\). Thus eventually \(|Q(h_n)-L|<\varepsilon\), which says \(Q(h_n)\to L\).
If the limit is not \(L\), there is some \(\varepsilon_0>0\) such that no matter how small \(\delta>0\) is, an admissible \(h\) exists with \(0<|h|<\delta\) and \(|Q(h)-L|\ge\varepsilon_0\).
For each \(n\in\mathbb{N}\), use \(\delta=1/n\) to choose such an increment \(h_n\). Then \(0<|h_n|<1/n\), so \(h_n\to0\), but \(|Q(h_n)-L|\ge\varepsilon_0\) for every \(n\). Therefore \(Q(h_n)\) does not tend to \(L\), contradicting the assumption that every such sequence does.
This proves both directions. In particular, if two admissible sequences approach zero and their quotient limits differ, a finite difference-quotient limit cannot exist: the theorem would require both sequences to tend to the same \(L\).
Worked Examples
Worked Example: The Absolute-Value Function
Let \(f(x)=|x|\) on \(\mathbb{R}\), and test differentiability at \(a=0\). First \(f(0)=0\), so for \(h\ne0\), the difference quotient is
Choose the positive sequence \(h_n=1/n\). Each term is nonzero, \(h_n\to0\), and
Now choose the negative sequence \(k_n=-1/n\). It also consists of nonzero increments and tends to zero. Since \(|-1/n|=1/n\),
The first quotient sequence tends to \(1\), while the second tends to \(-1\). Since these limits differ, the difference quotient has no limit at zero, and \(f\) is not differentiable there. Both sequences are admissible because the domain is all of \(\mathbb{R}\).
Worked Example: A Function Defined Differently on Rational and Irrational Inputs
Define \(f:\mathbb{R}\to\mathbb{R}\) by \(f(0)=0\), with \(f(x)=x\) for nonzero rational \(x\), and \(f(x)=2x\) for irrational \(x\). We test the derivative at zero. For every nonzero \(h\), \(Q(h)=f(h)/h\).
Take \(h_n=1/n\). Each \(h_n\) is a nonzero rational number, so
For a second sequence, take \(k_n=\sqrt{2}/n\). Each \(k_n\) is irrational: if \(\sqrt{2}/n\) were rational for a positive integer \(n\), then multiplying by \(n\) would make \(\sqrt{2}\) rational. Also \(k_n\ne0\) and \(k_n\to0\). Therefore
The quotient sequences are constant, so their limits are \(1\) and \(2\), respectively. They are incompatible. Hence \(f\) is not differentiable at zero. The conclusion follows from the quotient limits, not merely from the fact that the formula changes according to whether the input is rational or irrational.
Worked Example: A Finite Quotient Limit on One Side and Unbounded Quotients on the Other
Define \(f:\mathbb{R}\to\mathbb{R}\) by \(f(x)=\sqrt{x}\) for \(x>0\) and \(f(x)=0\) for \(x\le0\). At \(a=0\), we have \(f(0)=0\). For negative increments \(h<0\),
Thus the admissible sequence \(k_n=-1/n\) tends to zero and satisfies \(Q(k_n)=0\) for every \(n\), so \(Q(k_n)\to0\).
For positive increments \(h>0\), the quotient is
Take \(h_n=1/n^2\). These are positive admissible increments and \(h_n\to0\). Their quotients are
As \(n\to\infty\), \(Q(h_n)=n\) is unbounded and tends to \(+\infty\), so it does not tend to any finite real number. This conflicts with the negative sequence, whose quotients tend to zero. Therefore \(f\) has no finite derivative at zero.
How to Choose Useful Sequences
A sequence is useful when it simplifies the quotient or isolates a particular behavior. The examples suggest several practical choices:
- For a possible difference between left and right behavior, try \(h_n=1/n\) and \(k_n=-1/n\), when both are admissible.
- For a piecewise function, choose increments that remain in different pieces.
- For formulas involving special input sets, choose sequences that stay within those sets and verify that every term has the intended property.
- If the quotient may become unbounded, select increments that make its size explicit and check that the resulting quotient cannot converge to a finite value.
Always check the domain condition. A sequence of numbers tending to zero is not a valid test if one or more of its terms make \(a+h_n\) fall outside \(A\). Also verify that each increment is nonzero, because the difference quotient is undefined at \(h=0\).
A Common Mistake: Testing Only One Sequence
Suppose someone tests \(f(x)=|x|\) at zero using only \(h_n=1/n\), obtains \(Q(h_n)=1\), and concludes that \(f'(0)=1\). The calculation is correct, but the conclusion is not. It checks only one sequence, while differentiability requires the same limit for every admissible sequence approaching zero.
The correction is to test a second sequence from the other side. With \(k_n=-1/n\), the calculation gives \(Q(k_n)=-1\). Since \(1\ne-1\), the limit does not exist. One sequence can disprove a proposed value if its quotient fails to approach that value; however, one sequence whose quotient converges cannot establish that a derivative exists.
Check Your Understanding
For each question, verify that any sequence you use consists of admissible nonzero increments and tends to zero.
- Why must the point be an accumulation point of the domain, apart from itself, for the derivative limit to determine a value?
- If two admissible sequences of increments approach zero and their quotient limits are \(3\) and \(5\), what does this imply about the derivative?
- For \(f(x)=|x|\) at zero, calculate the difference quotient along \(h_n=-1/n\).
- Why does finding one sequence whose quotient tends to \(L\) not prove that the derivative exists and equals \(L\)?
- For a function with a restricted domain, what must be checked before a proposed sequence of increments can be used?
