When One Application Is Not Enough
A quotient may approach the form \(0/0\) even though its numerator and denominator both vanish rapidly near the same point. One application of L’Hôpital’s Rule may still leave a \(0/0\) quotient. In that situation, further applications can reveal which function vanishes faster and whether the original quotient has a finite limit.
The key is to keep track of the successive derivatives. If the first nonzero derivatives of the numerator and denominator occur at the same order, their ratio determines the limit, provided L’Hôpital’s Rule is valid at each step. If those first nonzero derivatives occur at different orders, comparing the orders can show that the quotient tends to zero or becomes unbounded.
For example, \(f(x)=x^3\) has a zero of order \(3\) at \(0\): its value, first derivative, and second derivative vanish there, while its third derivative is \(6\). The order records how many successive derivatives vanish before the first nonzero one.
Repeated L’Hôpital’s Rule
For a finite endpoint \(a\), the relevant one-step version of L’Hôpital’s Rule says the following: if \(u(x)\) and \(v(x)\) both tend to zero as \(x\to a\), the functions are differentiable on a punctured interval around \(a\), \(v'(x)\ne 0\) there, and \(\lim_{x\to a}u'(x)/v'(x)=L\), then \(\lim_{x\to a}u(x)/v(x)=L\). The limit may be one-sided if the domain or the interval under consideration is one-sided.
To apply the rule repeatedly, the same requirements must be checked at each application. In particular, before an application, the current numerator and denominator must both tend to zero (or another form covered by L’Hôpital’s Rule), and the derivative of the current denominator must not vanish on the punctured interval. For the repeated zero-over-zero setting here, it is enough to require that each denominator derivative used in the process is nonzero near \(a\), apart from possibly at \(a\) itself.
Here \(f^{(0)}\) means \(f\) itself. The nonvanishing condition ensures that each denominator derivative used in the repeated applications is allowed. The endpoint conditions ensure that, before the final application, both functions at the current stage still tend to zero.
Why the First Nonzero Derivatives Determine the Limit
The theorem follows by applying the one-step rule in succession. This gives a short proof, and it also explains why the first nonzero derivatives matter.
Since \(f(a)=g(a)=0\), both functions tend to zero at the endpoint. The one-step rule gives the limit of \(f/g\), if it exists, by examining \(f'/g'\).
For each \(j<r\), the assumptions give \(f^{(j)}(a)=g^{(j)}(a)=0\). Thus, after differentiating \(j\) times, the current quotient is still of type \(0/0\), and another application is available under the nonvanishing conditions.
After \(r\) applications, the quotient of derivatives has limit \(f^{(r)}(a)/g^{(r)}(a)\), by continuity and \(g^{(r)}(a)\ne 0\). Each earlier application preserves the limit, so this is also the limit of the original quotient.
This argument does not say to differentiate a quotient as though the derivative of \(f/g\) were \(f'/g'\). L’Hôpital’s Rule is a theorem relating two limits under specific hypotheses; the successive quotients are \(f'/g'\), \(f''/g''\), and so on.
Worked Example: Matching Third-Order Zeros
Worked Example: Matching Third-Order Zeros
Evaluate \(\displaystyle\lim_{x\to 0}\frac{e^x-1-x-\frac{x^2}{2}}{\sin x-x}\).
At \(x=0\), the numerator is \(1-1-0-0=0\), and the denominator is \(0-\sin 0=0\). Apply L’Hôpital’s Rule. The first derivatives are \(e^x-1-x\) and \(\cos x-1\), which both equal \(0\) at \(x=0\):
Apply the rule again. The second derivatives are \(e^x-1\) and \(-\sin x\), which again both equal \(0\) at \(0\):
A third application gives:
Thus the original limit is \(-1\). As a check on the derivative chain, the third derivatives are \(e^x\) and \(-\cos x\), whose values at \(0\) are \(1\) and \(-1\), respectively. Their ratio confirms the result.
When the Numerator Vanishes to Higher Order
If the denominator reaches a nonzero derivative before the numerator does, the quotient can tend to zero. Repeated applications are used only while both derivatives still give a zero-over-zero form. Once the denominator derivative is nonzero at the endpoint while the numerator derivative is zero there, continuity gives a zero limit for that final derivative quotient.
Worked Example: A Higher-Order Numerator
Evaluate \(\displaystyle\lim_{x\to 0}\frac{1-\cos x-\frac{x^2}{2}}{x-\sin x}\).
Both numerator and denominator equal \(0\) at \(x=0\). Their first derivatives, \(\sin x-x\) and \(1-\cos x\), also both vanish at \(0\). The first application therefore gives:
The second derivatives are \(\cos x-1\) and \(\sin x\), both zero at \(0\), so apply the rule again:
The third derivatives are \(-\sin x\) and \(\cos x\). The numerator tends to \(0\), while the denominator tends to \(1\), so the quotient tends to \(0/1=0\). No further application is needed: at this stage the form is not \(0/0\).
The answer is \(0\). The derivative check agrees: the numerator’s third derivative at \(0\) is \(0\), and the denominator’s third derivative is \(1\), so the derivative quotient has limit \(0\).
When the Numerator Vanishes to Lower Order
If the numerator reaches its first nonzero derivative sooner than the denominator, reversing the quotient can be useful. When the reversed quotient satisfies the repeated-rule conditions and tends to zero, the original quotient cannot have a finite limit; its magnitude becomes unbounded. The sign of the original quotient must still be determined from the functions on the side being considered.
Worked Example: Lower-Order Numerator and One-Sided Divergence
Evaluate \(\displaystyle\lim_{x\to 0^+}\frac{x-\sin x}{1-\cos x-\frac{x^2}{2}}\).
Both functions vanish at \(0\). Consider the reciprocal quotient. Its first two derivatives are \(\sin x-x\) and \(1-\cos x\) in the numerator, and \(1-\cos x\) and \(\sin x\) in the denominator. These pairs both vanish at \(0\), so L’Hôpital’s Rule applies twice:
One more application gives:
For \(x>0\) sufficiently close to \(0\), \(x-\sin x>0\), while \(1-\cos x-\frac{x^2}{2}<0\). For example, these signs follow from the local leading terms \(x-\sin x\sim x^3/6\) and \(1-\cos x-\frac{x^2}{2}\sim -x^4/24\). Thus the reciprocal quotient tends to \(0\) through negative values, and its reciprocal tends to \(-\infty\). Therefore the requested limit is \(-\infty\).
Checking the derivative calculation at the final stage gives numerator derivative \(-\sin x\) and denominator derivative \(\cos x\), with endpoint values \(0\) and \(1\). This confirms that the reciprocal limit is zero; the sign check determines the direction of divergence for the original quotient.
A Common Mistake: Continuing Without Checking the Form
Repeated differentiation is not automatic. Each use of L’Hôpital’s Rule requires a qualifying indeterminate form at that stage. For example, in the second worked example, after two applications the quotient is \((\cos x-1)/\sin x\), still \(0/0\), so a third application is valid. After that, the quotient is \(-\sin x/\cos x\), which tends to \(0/1\), not \(0/0\). Applying the rule again would not be justified.
A related error is to report the ratio of higher derivatives without establishing that the intervening applications are allowed. The derivatives at the endpoint alone do not replace the hypotheses of L’Hôpital’s Rule: the relevant functions must be differentiable on a punctured interval, each denominator derivative used must be nonzero there, and each current quotient must have the required form.
Check Your Understanding
For each question, identify the vanishing orders and state which conditions you would check before applying L’Hôpital’s Rule repeatedly.
- If \(f(a)=g(a)=0\), \(f'(a)=g'(a)=0\), and \(g''(a)\ne0\), what additional endpoint information would let the second derivatives determine the quotient limit?
- Why must the current numerator and denominator still tend to zero before another zero-over-zero application?
- If the numerator’s first nonzero derivative is of higher order than the denominator’s, what outcome can occur, as in the second worked example?
- Why is it useful to examine the reciprocal when the numerator vanishes to lower order?
- In the first worked example, what are the values at zero of the third derivatives of the numerator and denominator?
