What the Converse Would Claim
A limit of a quotient and a limit of a derivative quotient are different statements. L'Hôpital's Rule gives a useful implication in one direction: under its hypotheses, a limit of the derivative quotient can establish the limit of the original quotient. It does not say that the derivative quotient must have a limit whenever the original quotient does.
To make the logical distinction precise, consider a finite endpoint \(a\) and functions \(f\) and \(g\) that both approach zero there. A standard \(0/0\) form of L'Hôpital's Rule says that if \(f\) and \(g\) are differentiable near \(a\), \(g'\) does not vanish on the punctured interval under consideration, and
then, subject to those hypotheses,
The proposed converse would reverse this implication: if \(f(x)/g(x)\) has a limit, then \(f'(x)/g'(x)\) must have a limit, perhaps the same one. That claim is false. The examples below meet the relevant \(0/0\) setup and have \(g'(x)=1\), yet their derivative quotients fail to converge.
Worked Example: A Zero Limit with Bounded Oscillation
Worked Example: A Zero Limit with Bounded Oscillation
For \(x>0\), let \(f(x)=x^2\sin(1/x)\) and \(g(x)=x\), and set \(f(0)=g(0)=0\). Find the limit of \(f(x)/g(x)\) as \(x\to0^+\), and determine whether \(f'(x)/g'(x)\) has a limit.
First simplify the original quotient for \(x>0\):
Since \(|\sin(1/x)|\leq 1\), we have
As \(x\to0^+\), the right-hand side tends to zero, so the squeeze theorem gives \(\lim_{x\to0^+}f(x)/g(x)=0\).
Now differentiate for \(x>0\). By the product rule and chain rule,
Thus \(f'(x)/g'(x)=2x\sin(1/x)-\cos(1/x)\). To test whether this has a limit, take the two sequences
Both sequences approach zero through positive values. At \(x_n\), \(\sin(1/x_n)=0\) and \(\cos(1/x_n)=1\), so the derivative quotient is \(-1\). At \(y_n\), \(\sin(1/y_n)=0\) and \(\cos(1/y_n)=-1\), so it is \(1\). Since the values along these two sequences have different limits, \(f'(x)/g'(x)\) has no limit as \(x\to0^+\).
As a check on the endpoint setup, \(f(0)=g(0)=0\), and \(g'(x)=1\) for \(x>0\). Also, the difference quotient for \(f\) at zero is \(f(x)/x=x\sin(1/x)\), which tends to zero, so \(f'(0)=0\). The original quotient nevertheless converges even though the derivative quotient does not.
Why Sequences Prove Nonexistence of a Limit
A function with a limit \(L\) at an endpoint must approach \(L\) along every sequence in its domain that approaches that endpoint. This gives a direct way to prove that a limit does not exist: find two such sequences along which the function approaches different values. In the example, the derivative quotient takes the constant value \(-1\) along one sequence and \(1\) along another. It therefore cannot have a single limit.
The quotient limit was established in a different way. Its absolute value is bounded above by \(x\), which tends to zero. This contrast is useful: the two limits involve related functions, but each must be checked on its own. A limit of \(f/g\) does not impose the same point-by-point behavior on \(f'/g'\).
Worked Example: A Nonzero Limit with Oscillation
Worked Example: A Nonzero Limit with Oscillation
For \(x>0\), let \(f(x)=x+x^2\sin(1/x)\) and \(g(x)=x\), and set \(f(0)=g(0)=0\). This time the quotient has a nonzero limit.
For \(x>0\), divide each term in \(f(x)\) by \(x\):
Again, \(|\sin(1/x)|\leq1\), so \(|x\sin(1/x)|\leq x\to0\). Therefore
Differentiate \(f\) term by term, using the product rule on the second term:
Hence \(f'(x)/g'(x)=1+2x\sin(1/x)-\cos(1/x)\). Use the same sequences \(x_n=1/(2\pi n)\) and \(y_n=1/((2n+1)\pi)\). At \(x_n\), the sine term is zero and the cosine is \(1\), giving derivative quotient \(0\). At \(y_n\), the sine term is zero and the cosine is \(-1\), giving derivative quotient \(2\). Both sequences tend to zero, but the resulting values differ, so the derivative quotient has no limit.
For completeness, \(f(x)/x=1+x\sin(1/x)\to1\), which is exactly the difference quotient for \(f\) at zero. Thus \(f'(0)=1\). The quotient itself tends to \(1\), while its derivative quotient oscillates between values approaching \(0\) and \(2\). This also rules out a converse that would require the derivative quotient to tend to the same value as the original quotient.
Why the Two Quotients Can Behave Differently
The original quotient compares the values \(f(x)\) and \(g(x)\) at each point. The derivative quotient compares their instantaneous rates of change. These are related quantities, but they are not interchangeable. In particular, a small correction to the values of \(f\) can disappear after division by \(g\) as \(x\) approaches the endpoint, while differentiating that correction can reveal an oscillating term.
For example, in the preceding calculation the correction \(x^2\sin(1/x)\), after division by \(x\), becomes \(x\sin(1/x)\), which tends to zero. Its derivative contains \(-\cos(1/x)\), which continues to oscillate. The quotient limit is controlled by the size of the correction in the original functions; the derivative quotient responds to how rapidly that correction varies.
This explanation is not a new test for limits. It describes why there is no logical contradiction when a quotient converges but its derivative quotient does not. L'Hôpital's Rule can still be applied when its conditions and derivative-quotient limit are available. It simply cannot be reversed as a general implication.
Worked Example: An Unbounded Derivative Quotient
Worked Example: An Unbounded Derivative Quotient
For \(x>0\), let \(f(x)=x^2\sin(1/x^2)\) and \(g(x)=x\), with \(f(0)=g(0)=0\). This example shows that the derivative quotient can fail to have a limit by becoming unbounded along different sequences.
The original quotient simplifies to
Its absolute value satisfies
Since \(x\to0^+\), the squeeze theorem gives \(\lim_{x\to0^+}f(x)/g(x)=0\).
By the chain rule,
Now define positive sequences
Both sequences tend to zero. At either sequence, \(\sin(1/x^2)=0\). At \(u_n\), \(\cos(1/u_n^2)=1\), giving \(f'(u_n)/g'(u_n)=-2/u_n\), which tends to \(-\infty\). At \(v_n\), \(\cos(1/v_n^2)=-1\), giving \(f'(v_n)/g'(v_n)=2/v_n\), which tends to \(+\infty\). The derivative quotient therefore has no limit. Meanwhile, the original quotient tends to zero as already checked.
Finally, \(f(x)/x=x\sin(1/x^2)\to0\), so \(f'(0)=0\). As in the previous examples, the functions meet the \(0/0\) endpoint setup and \(g'(x)=1\) for \(x>0\). The failure is in the proposed converse, not in the existence of the original quotient limit.
The Logical Lesson
These examples establish three distinct cases:
| Limit of \(f/g\) | Behavior of \(f'/g'\) | What it demonstrates |
|---|---|---|
| Zero | Bounded oscillation | The derivative quotient need not converge to the quotient limit. |
| One | Bounded oscillation | The failure also occurs for a nonzero quotient limit. |
| Zero | Unbounded oscillation | The derivative quotient need not even remain bounded. |
When using L'Hôpital's Rule, a derivative quotient with a limit can be a powerful route to the original limit. But if that derivative quotient has no limit, the correct conclusion is only that this particular route does not establish the answer. The original quotient may still have a limit, as each example shows.
The key distinction is between a sufficient condition and a necessary condition. A sufficient condition guarantees a conclusion when it holds. It need not hold in every case where that conclusion is true. L'Hôpital's Rule supplies a sufficient condition for evaluating certain quotient limits; its derivative-quotient condition is not necessary for the original quotient to converge.
Check Your Understanding
Use the examples and the logical distinction above to answer these questions.
- What implication does L'Hôpital's Rule give in the \(0/0\) case, and what would the false converse claim?
- Why do two sequences that give different limiting values for \(f'(x)/g'(x)\) show that this derivative quotient has no limit?
- In the example \(f(x)=x^2\sin(1/x)\), \(g(x)=x\), how does the squeeze theorem establish the limit of \(f(x)/g(x)\)?
- Give one reason that the derivative quotient can oscillate even when the original quotient converges.
- If the derivative quotient does not have a limit, what can you conclude about the original quotient from that fact alone?
