Why Checking the Hypotheses Matters
L’Hôpital’s Rule can turn a difficult quotient limit into a simpler limit of derivatives. But it is not a general instruction to differentiate the numerator and denominator whenever substitution produces a difficult expression. The rule is a theorem with hypotheses, and each hypothesis must hold near the point being approached.
This tutorial focuses on a finite endpoint \(c\). The functions may be defined only on one side of \(c\), so it is useful to name the interval on which they are being considered. For example, the interval may be \((c,c+\delta)\) or \((c-\delta,c)\), where \(\delta>0\). The rule concerns behavior as \(x\) approaches \(c\) within that interval.
The essential checks are local: the numerator and denominator must have the required limiting behavior, both functions must be differentiable on the punctured interval, and the denominator’s derivative must not vanish there. A conclusion about the derivative quotient is useful only after these conditions have been verified.
The Conditions Near a Finite Endpoint
In the first case, \(f/g\) has the indeterminate form \(0/0\). In the second, it has the form \(\infty/\infty\), where either function may tend to positive or negative infinity. The conditions concern limits, so the values \(f(c)\) and \(g(c)\) need not be defined.
Differentiability is required on a whole punctured interval near \(c\), not merely at a few points or along a sequence. The condition \(g'(x)\ne0\) is also required throughout some such interval. It is not enough that \(g'\) is nonzero at most nearby points, or just that its derivative quotient has a limit where that quotient happens to be defined.
The phrase “throughout some such interval” is important. A derivative might vanish farther away from \(c\), without affecting the rule: the interval can be shortened. But if zeros of \(g'\) occur arbitrarily close to \(c\), no sufficiently small punctured interval avoids them, and this version of the rule does not apply.
A Proof in the Zero-Over-Zero Case
The zero-over-zero version is a direct consequence of the Cauchy Mean Value Theorem. The proof also explains why the derivative quotient appears. Suppose \(f(x)\to0\) and \(g(x)\to0\) as \(x\to c\) within the chosen interval, and that \(f\) and \(g\) are differentiable there with \(g'(x)\ne0\).
Define \(f(c)=0\) and \(g(c)=0\) for the purpose of this argument. The limiting assumptions make these extensions continuous at \(c\). For a nearby \(x\ne c\), the Cauchy Mean Value Theorem applied between \(c\) and \(x\) gives a point \(\xi\) strictly between them such that
The denominator on the left is nonzero. To see this, if \(g(x)=g(c)=0\) for some \(x\ne c\), Rolle’s Theorem would give a point between \(c\) and \(x\) where \(g'\) is zero, contrary to the hypothesis. Since \(f(c)=g(c)=0\), the equality becomes
As \(x\) approaches \(c\), the intermediate point \(\xi\) lies between \(x\) and \(c\), so it also approaches \(c\). Therefore, if \(f'(t)/g'(t)\to L\) as \(t\to c\), the right-hand side approaches \(L\), and so does \(f(x)/g(x)\). The same reasoning applies if the derivative quotient tends to \(+\infty\) or \(-\infty\). This proves the zero-over-zero conclusion.
This argument uses the Cauchy Mean Value Theorem on an interval ending at \(c\), with continuity at the endpoint supplied by the zero limits. The infinity-over-infinity case is another standard form of L’Hôpital’s Rule; its hypotheses must still be checked separately rather than inferred from the zero-over-zero proof.
Worked Examples: Checking Before Differentiating
Worked Example: A Zero-Over-Zero Limit at a Finite Endpoint
Evaluate
First check substitution. The numerator tends to \(2^2-4=0\), and the denominator tends to \(3(2)-6=0\), so the form is \(0/0\). Both functions are polynomials, hence differentiable on a punctured interval around \(2\). The denominator is \(g(x)=3x-6\), with derivative \(g'(x)=3\), which is nonzero throughout that interval. The hypotheses are satisfied.
Differentiate the numerator and denominator separately:
The derivative quotient has limit
Thus L’Hôpital’s Rule gives \(4/3\). As a check, for \(x\ne2\),
Worked Example: An Infinity-Over-Infinity Limit at a Finite Endpoint
Evaluate the limit from the right:
On a sufficiently small interval \((0,\delta)\), both functions are differentiable. As \(x\to0^+\), the numerator \(1/x\) tends to \(+\infty\), and the denominator \(2/x+1\) also tends to \(+\infty\). Thus the quotient has the form \(\infty/\infty\).
Write \(f(x)=1/x\) and \(g(x)=2/x+1\). Their derivatives are
For every \(x\in(0,\delta)\), \(g'(x)=-2/x^2\ne0\). The derivative quotient is
Its limit is \(1/2\), so L’Hôpital’s Rule gives the original limit as \(1/2\). Check directly by multiplying numerator and denominator by \(x\), which is positive and nonzero on this interval:
Worked Example: When the Denominator Derivative Vanishes Nearby
Worked Example: A Rule That Cannot Be Applied on Any Punctured Interval
Consider, for \(x>0\),
Since \(1\le 2+\sin(1/x)\le3\), the factor in parentheses is bounded, and \(x^2\to0\). Thus \(f(x)\to0\) and \(g(x)\to0\). The functions are differentiable for every \(x>0\).
Differentiate \(g\) using the product and chain rules:
For positive integers \(n\), define \(a_n=1/(2\pi n)\) and \(b_n=1/((2n+1)\pi)\). At both points the sine term is zero. At \(a_n\), the cosine is \(1\), while at \(b_n\) it is \(-1\). Therefore,
For all sufficiently large \(n\), \(4a_n<1\), so \(g'(a_n)<0\); also \(g'(b_n)>0\). The derivative \(g'\) is continuous between \(a_n\) and \(b_n\), since that interval does not contain zero. The Intermediate Value Theorem therefore gives a point between them where \(g'\) is zero. These intervals approach zero as \(n\) increases, so such zeros occur arbitrarily close to the endpoint.
Consequently, there is no punctured interval \((0,\delta)\) on which \(g'\) is everywhere nonzero. Even though \(f'=g'\), it is not valid to apply L’Hôpital’s Rule here: the derivative quotient is undefined where \(g'=0\), and the required condition fails arbitrarily close to zero.
The limit itself is easy to check without the rule. For every \(x>0\), the common factor is positive, so
The correction is not to ignore the zeros of \(g'\), but to use this direct identity. A limit can exist even when a particular theorem does not apply.
A Practical Hypothesis Check
Before using the rule near a finite endpoint, check the conditions in order. This prevents a familiar-looking quotient from being treated as though the conclusion were automatic.
Identify the finite endpoint \(c\) and the side or interval on which \(x\) approaches it.
Substitute the limiting behavior of the numerator and denominator. The quotient must be \(0/0\) or \(\infty/\infty\).
Both functions must be differentiable throughout a punctured interval near \(c\).
Verify \(g'(x)\ne0\) at every point of some sufficiently small punctured interval, not just at selected points.
If it exists as a finite value or signed infinity, the rule gives the limit of the original quotient.
A Common Mistake
A frequent error is to write “the form is \(0/0\), so apply L’Hôpital’s Rule.” The form is only one condition. For instance, in the third example, both functions tend to zero, but the denominator derivative vanishes arbitrarily close to the endpoint. The rule’s hypotheses are not met, so the derivative quotient cannot be used to establish the original limit.
Another important distinction is that L’Hôpital’s Rule concerns the quotient \(f'(x)/g'(x)\), not the derivative of the quotient \(f(x)/g(x)\). The theorem does not say that differentiating a fraction produces the desired limit. It says that under its hypotheses, a limit of one particular derivative quotient determines the limit of the original quotient.
Why the Local Conditions Matter
The hypotheses are local because a limit depends on values arbitrarily close to the endpoint. A condition that holds only at \(c\) would not be enough: the rule compares function values through derivatives at nearby points. Likewise, checking that \(g'\) is nonzero at a few sample points cannot rule out zeros between them or zeros accumulating at \(c\).
The derivative condition also protects the quotient \(f'/g'\) from having a zero denominator on the interval where the argument is used. In the zero-over-zero proof, it ensures that the Cauchy Mean Value Theorem gives a valid comparison and that \(g(x)\ne g(c)\) for nearby \(x\ne c\). The theorem’s assumptions are part of the reasoning, not formalities to skip.
Summary
For L’Hôpital’s Rule near a finite endpoint, first establish the correct indeterminate form, then verify differentiability on a punctured interval and the nonvanishing of the denominator derivative throughout a sufficiently small such interval. Only after these checks should the derivative quotient be evaluated. In the zero-over-zero case, the Cauchy Mean Value Theorem explains why that quotient controls the original one.
Check Your Understanding
For each question, focus on which hypothesis is being checked and why it matters.
- What two limiting forms permit the standard versions of L’Hôpital’s Rule stated here?
- Why must differentiability hold on a punctured interval, rather than only at the endpoint?
- If \(g'\) has zeros arbitrarily close to \(c\), can the interval be shortened to make the stated nonvanishing condition hold?
- In the zero-over-zero proof, why does \(g(x)\ne g(c)\) for nearby \(x\ne c\)?
- If \(f'(x)/g'(x)\) has a limit but the original quotient is not in one of the required forms, may the rule be applied as stated?
