A Quotient Can Converge While Its Derivative Ratio Does Not
L'Hôpital's Rule is a powerful way to evaluate certain limits of quotients. It relates the limit of a quotient to the limit of the ratio of the derivatives of its numerator and denominator. That relationship is useful, but it is not an equivalence: the original quotient can have a limit even when the derivative ratio does not.
The functions in this example make that distinction visible. Their quotient approaches zero as \(x\) approaches zero, while the derivative ratio keeps taking values near both \(1\) and \(-1\). The quotient limit follows from a direct bound, and the failure of the derivative ratio to converge follows by examining two sequences.
The Functions and Their Quotient
For \(x\ne 0\), let
Both functions approach zero as \(x\to 0\), so their quotient initially has the indeterminate form \(0/0\). For every \(x\ne 0\), the quotient simplifies to
The sine factor oscillates increasingly rapidly near zero. But it is always between \(-1\) and \(1\). That bound controls the whole product: the factor \(x\) shrinks to zero, regardless of how the sine factor varies.
Worked Example: Evaluate the Original Quotient Limit
Worked Example: Evaluate the Original Quotient Limit
Find \(\lim_{x\to 0} f(x)/g(x)\) for the functions above.
For \(x\ne 0\), the quotient is \(x\sin(1/x)\). Since \(-1\le \sin(1/x)\le 1\), multiplication by \(x\) requires care about the sign of \(x\). An absolute-value bound avoids that issue:
The lower and upper bounds can therefore be written as
As \(x\to 0\), both \(-|x|\) and \(|x|\) tend to zero. The squeeze theorem gives
Check the bound: the absolute value of the quotient is at most \(|x|\), which can be made arbitrarily small by taking \(x\) sufficiently close to zero. Thus the value zero is consistent with, and forced by, the estimate.
Compute the Derivative Ratio
For \(x\ne 0\), differentiate \(f(x)=x^2\sin(1/x)\) using the product rule and chain rule. The derivative of \(x^2\) is \(2x\), and the derivative of \(\sin(1/x)\) is \(\cos(1/x)(-1/x^2)\). Hence
Since \(g(x)=x\), its derivative is \(g'(x)=1\). Therefore the derivative ratio, for \(x\ne 0\), is
The first term, \(2x\sin(1/x)\), tends to zero: its absolute value is at most \(2|x|\). The cosine term, however, does not approach a single value. We can establish the resulting failure of the derivative ratio to converge precisely by choosing sequences.
Worked Example: Find Two Sequences with Different Derivative-Ratio Values
For positive integers \(n\), define
Both sequences consist of positive numbers and approach zero as \(n\to\infty\). At \(x=a_n\), the reciprocal is \(1/a_n=2\pi n\). Thus \(\sin(1/a_n)=0\), \(\cos(1/a_n)=1\), and
At \(x=b_n\), the reciprocal is \(1/b_n=(2n+1)\pi\). Therefore \(\sin(1/b_n)=0\), \(\cos(1/b_n)=-1\), and
The derivative ratio equals \(-1\) along a sequence approaching zero and \(1\) along another sequence approaching zero. If it had a limit at zero, its values along every sequence approaching zero would tend to that same limit. These two constant sequences of values contradict that requirement. Thus the derivative ratio has no limit.
Check: \(a_n\to0\) and \(b_n\to0\), while the corresponding ratios remain exactly \(-1\) and \(1\). The difference is not a small approximation error; the two subsequential values are distinct.
Why the Two Conclusions Are Compatible
The original quotient and the derivative ratio are different functions. In this case, the original quotient is \(x\sin(1/x)\), whose absolute value is bounded by \(|x|\). The derivative ratio is \(2x\sin(1/x)-\cos(1/x)\), which includes an unshrinking oscillatory term.
The distinction is captured by the direction of L'Hôpital's Rule. In its \(0/0\) form, a standard version says that if \(f\) and \(g\) are differentiable on a punctured interval around \(a\), \(f(x)\to 0\) and \(g(x)\to 0\) as \(x\to a\), \(g'(x)\ne 0\) there, and \(f'(x)/g'(x)\) tends to a finite limit \(L\), then \(f(x)/g(x)\) tends to \(L\), provided the quotient is defined on that punctured interval. The rule gives a conclusion when its derivative-ratio limit exists and its hypotheses hold.
Here, the \(0/0\) conditions on \(f\) and \(g\) hold, both functions are differentiable for \(x\ne 0\), and \(g'(x)=1\ne 0\). But the derivative ratio has no limit. So this version of L'Hôpital's Rule does not produce a conclusion about the quotient. It does not follow that the quotient has no limit; the direct bound already proves that it does.
Worked Example: Check the Quotient Limit Along the Same Sequences
Worked Example: Check the Quotient Limit Along the Same Sequences
Use \(a_n=1/(2\pi n)\) and \(b_n=1/((2n+1)\pi)\) to evaluate the original quotient along each sequence.
Since \(\sin(1/a_n)=\sin(2\pi n)=0\),
Similarly, \(\sin(1/b_n)=\sin((2n+1)\pi)=0\), so
These particular sequences do not by themselves prove that the quotient tends to zero: checking selected sequences is not a substitute for controlling all \(x\) near zero. The earlier bound does provide that control. Here, the calculation confirms that the two sequences producing different derivative-ratio values do not produce different quotient limits.
Check: both sequences give quotient value zero for every positive integer \(n\), in agreement with the limit established by the squeeze theorem.
A Common Mistake: Reversing the Rule
A tempting but invalid argument is: “The derivative ratio has no limit, so the original quotient has no limit.” This reverses the implication supplied by L'Hôpital's Rule. A rule that can prove a quotient limit when the derivative ratio converges does not say that the quotient must fail to converge when the derivative ratio does not.
For these functions, the incorrect inference would stop after observing the values \(-1\) and \(1\) along the two sequences and conclude that \(f(x)/g(x)\) has no limit. The correction is to examine \(f(x)/g(x)\) itself:
That bound proves the quotient limit directly. The derivative ratio is useful information about the derivatives, but its oscillation does not undo the bound on the original quotient.
Why This Example Matters
This example clarifies what to do when applying L'Hôpital's Rule. First verify that the original quotient has an appropriate indeterminate form and that the differentiability and denominator-derivative conditions hold. Then evaluate the derivative ratio only if its limit can help. If that ratio does not have a limit, the method has not resolved the original limit; another argument may still succeed.
In particular, a direct inequality can be more informative than differentiating. The factor \(x\) in \(x\sin(1/x)\) forces the product toward zero because the sine factor remains bounded. In the derivative, differentiating the rapidly varying input \(1/x\) introduces a term that retains the oscillation. The two behaviors are compatible because the derivative ratio is not the original quotient.
Check Your Understanding
Answer each question using the functions \(f(x)=x^2\sin(1/x)\) and \(g(x)=x\) near zero.
- What inequality bounds the absolute value of \(f(x)/g(x)\), and how does it establish the quotient limit?
- Compute \(f'(x)\) for \(x\ne0\), showing where the chain rule contributes a factor of \(-1/x^2\).
- What values does \(f'(x)/g'(x)\) take along \(a_n=1/(2\pi n)\) and \(b_n=1/((2n+1)\pi)\)?
- Why do those two sequences prove that the derivative ratio has no limit?
- Why does the failure of the derivative ratio to converge not imply that the original quotient has no limit?
