Why a Local Condition Matters
L'Hôpital's Rule compares the behavior of two functions near an endpoint by comparing their derivatives nearby. That comparison is only valid when the denominator derivative is nonzero throughout a sufficiently small punctured neighborhood of the endpoint. The word throughout matters: it is not enough that the derivative be nonzero at most nearby points, or that a derivative quotient have a limit wherever it happens to be defined.
The condition concerns points close to the endpoint, not necessarily the endpoint itself. Nor does a zero farther away automatically prevent use of the rule: the interval can sometimes be made smaller. We will separate these cases and see why zeros accumulating at the endpoint are different.
The Finite-Endpoint Rule and Its Hypotheses
Here is a right-hand version of L'Hôpital's Rule. Suppose \(f\) and \(g\) are continuous on \([a,b)\), differentiable on \((a,b)\), and \(f(a)=g(a)=0\). Suppose also that \(g'(x)\ne 0\) for every \(x\) in \((a,b)\), and that \(\lim_{x\to a^+} f'(x)/g'(x)=L\), where \(L\) is finite. Then \(\lim_{x\to a^+} f(x)/g(x)=L\).
The rule has versions for left-hand endpoints and for limits at infinity. In each case, its hypotheses must hold on the relevant punctured interval or tail. This tutorial focuses on the finite right-hand version, where the distinction between the endpoint and nearby points is especially visible.
Why the Condition Works
The Cauchy Mean Value Theorem explains the role of \(g'\). For any \(x\in(a,b)\), apply that theorem to \(f\) and \(g\) on \([a,x]\). There is a point \(c_x\in(a,x)\) such that \[ \frac{f(x)-f(a)}{g(x)-g(a)} = \frac{f'(c_x)}{g'(c_x)}. \] Since \(f(a)=g(a)=0\), and \(g'(c_x)\ne0\), the equality becomes \[ \frac{f(x)}{g(x)} = \frac{f'(c_x)}{g'(c_x)}. \] In particular, the nonzero-derivative condition also ensures \(g(x)\ne0\): by the Mean Value Theorem, \(g(x)-g(a)=g'(d)(x-a)\ne0\) for some \(d\in(a,x)\).
As \(x\to a^+\), the point \(c_x\), which lies between \(a\) and \(x\), also approaches \(a\). If \(f'(t)/g'(t)\to L\) as \(t\to a^+\), then \(f'(c_x)/g'(c_x)\to L\). The displayed equality gives the desired limit of \(f(x)/g(x)\). This proof needs the derivative quotient to be defined at every such \(c_x\); if \(g'\) vanishes arbitrarily close to \(a\), that guarantee is lost.
Worked Example: A Zero Derivative at the Endpoint Is Allowed
Worked Example: A Zero Derivative at the Endpoint Is Allowed
Evaluate \(\lim_{x\to0^+}x^3/x^2\) using L'Hôpital's Rule, and check the hypotheses.
Set \(f(x)=x^3\) and \(g(x)=x^2\). Both are continuous at \(0\), differentiable for \(x>0\), and \(f(0)=g(0)=0\), so the quotient has the required \(0/0\) form. Their derivatives are \[ f'(x)=3x^2,\qquad g'(x)=2x. \] Although \(g'(0)=0\), for every \(x>0\) we have \(g'(x)=2x>0\). Thus the derivative is nonzero throughout the punctured interval \((0,1)\). Applying the rule gives \[ \lim_{x\to0^+}\frac{x^3}{x^2} = \lim_{x\to0^+}\frac{3x^2}{2x} = \lim_{x\to0^+}\frac{3x}{2} =0. \]
Check directly: for \(x>0\), \(x^3/x^2=x\), whose limit as \(x\to0^+\) is \(0\). The result agrees. The zero of \(g'\) at \(0\) is not a problem because \(0\) is the endpoint, not a point in the punctured interval.
Worked Example: Zeros Accumulating at the Endpoint
Worked Example: Zeros Accumulating at the Endpoint
Consider \(f(x)=g(x)=x^2(2+\sin(1/x))\) for \(x>0\), with \(f(0)=g(0)=0\). The quotient is easy to evaluate, but the denominator derivative vanishes at points arbitrarily close to \(0\).
For every \(x>0\), \(2+\sin(1/x)\ge1\), so \(g(x)>0\). Also, \[ \lim_{x\to0^+}x^2(2+\sin(1/x))=0, \] because \(0\le x^2(2+\sin(1/x))\le3x^2\). Thus \(g\) is continuous at \(0\), as is \(f\), and both have value zero there. For \(x>0\), their quotient is \[ \frac{f(x)}{g(x)} = \frac{x^2(2+\sin(1/x))}{x^2(2+\sin(1/x))} =1. \] Therefore the quotient limit is \(1\).
Differentiate \(g\) for \(x>0\), using the product and chain rules: \[ g'(x)=2x(2+\sin(1/x))-\cos(1/x). \] To show there are zeros arbitrarily close to \(0\), set \(t=1/x\) and consider \(H(t)=2(2+\sin t)/t-\cos t\), so \(g'(1/t)=H(t)\). For positive integers \(n\), evaluate at \(t_n=\pi/2+2\pi n\) and \(s_n=7\pi/4+2\pi n\). At the first points, \(\sin t_n=1\) and \(\cos t_n=0\), giving \[ H(t_n)=\frac{6}{t_n}>0. \] At the second points, \(\sin s_n=-\sqrt{2}/2\) and \(\cos s_n=\sqrt{2}/2\), giving \[ H(s_n)=\frac{4-\sqrt{2}}{s_n}-\frac{\sqrt{2}}{2}. \] Since \(s_n\to\infty\), this expression is negative for all sufficiently large \(n\). By continuity of \(H\), for each such \(n\) there is a \(u_n\in(t_n,s_n)\) with \(H(u_n)=0\). Hence \(g'(1/u_n)=0\), and \(1/u_n\to0^+\).
So \(g'\) has zeros inside every interval \((0,b)\). The standard hypothesis \(g'(x)\ne0\) throughout some punctured neighborhood fails. Indeed, at each of these points \(f'=g'=0\), so \(f'/g'\) is not even defined there. The quotient's limit is still \(1\), but that limit follows from the direct cancellation, not from an application of L'Hôpital's Rule.
Worked Example: A Zero Outside the Neighborhood Can Be Excluded
Worked Example: A Zero Outside the Neighborhood Can Be Excluded
Let \(f(x)=x^2\) and \(g(x)=x^2(1-x)\), with endpoint \(a=0\). A zero of \(g'\) occurs at \(x=2/3\), but this does not stop us using the rule near \(0\).
For \(0<x<1\), \(g(x)>0\); also \(f(0)=g(0)=0\). Differentiating gives \[ f'(x)=2x,\qquad g'(x)=2x(1-x)-x^2=2x-3x^2=x(2-3x). \] The derivative \(g'\) vanishes at \(x=0\) and \(x=2/3\). Choose, for example, the neighborhood \(0<x<1/2\). There \(x>0\) and \(2-3x>2-3/2=1/2>0\), so \(g'(x)>0\) throughout that interval. L'Hôpital's Rule applies there: \[ \lim_{x\to0^+}\frac{x^2}{x^2(1-x)} = \lim_{x\to0^+}\frac{2x}{2x-3x^2} = \lim_{x\to0^+}\frac{2}{2-3x} =1. \]
Check directly: for \(x>0\), \(f(x)/g(x)=1/(1-x)\), which tends to \(1\) as \(x\to0^+\). The zero at \(2/3\) does not matter because it lies outside the chosen interval \((0,1/2)\). The rule requires a suitable neighborhood, not that \(g'\) be nonzero across the entire domain of the functions.
A Common Mistake
A tempting but invalid argument in the oscillatory example is: “Since \(f=g\), their derivatives are equal, so \(f'/g'=1\); L'Hôpital's Rule gives the limit \(1\).” The error is treating \(f'/g'\) as defined even where its denominator is zero. At every point in that example where \(g'=0\), we also have \(f'=0\), and the derivative quotient is \(0/0\), which is undefined. More importantly, such points occur arbitrarily close to the endpoint, so there is no punctured neighborhood on which the standard nonvanishing hypothesis holds.
The correction is to check the hypotheses before using the rule. If there are zeros of \(g'\) only outside a smaller interval, choose an interval that excludes them, as in the third example. If zeros occur in every punctured neighborhood, the standard theorem cannot be invoked on any such interval. The original quotient may still have a limit, but it must be established by another valid argument or by a version of a theorem whose hypotheses are actually satisfied.
What to Check Near the Endpoint
When applying the finite-endpoint form, check the denominator derivative locally rather than merely evaluating it at the endpoint. In practice, this means asking whether there exists a single interval next to the endpoint on which \(g'\) is defined and nowhere zero. A derivative that vanishes at the endpoint may be entirely acceptable; a sequence of zeros approaching the endpoint is not.
For a right-hand limit at \(a\), examine points \(x>a\) sufficiently close to \(a\).
Verify the required indeterminate form, continuity at the endpoint, and differentiability nearby for the theorem being used.
Find a \(b>a\) such that \(g'(x)\ne0\) for every \(x\in(a,b)\). Checking only a few sample points is not enough.
Then evaluate the derivative-quotient limit. If zeros of \(g'\) accumulate at \(a\), use another argument for the original quotient.
Why the Distinction Matters
The nonvanishing condition is what lets the Cauchy Mean Value Theorem relate the original quotient to a derivative quotient at an intermediate point. It also ensures that the denominator function itself does not vanish again inside the interval. If the derivative has zeros accumulating at the endpoint, that comparison cannot be guaranteed, even if a derivative quotient appears well behaved wherever it is defined.
Keep three situations distinct: a zero at the endpoint is allowed; a zero separated from the endpoint can often be excluded by shrinking the interval; and zeros arbitrarily close to the endpoint prevent the standard nonvanishing hypothesis from holding. This is a local condition, but it is a condition on every point in a sufficiently small punctured neighborhood.
Check Your Understanding
For each question, distinguish what happens at the endpoint from what happens nearby.
- In the finite right-hand version of L'Hôpital's Rule, may \(g'(a)=0\)? What must be true of \(g'\) for \(a<x<b\)?
- Why does \(f(x)=g(x)=x^2(2+\sin(1/x))\) have quotient limit \(1\), and why is the standard rule unavailable there?
- If \(g'\) vanishes at one point \(c>a\), what should you check before deciding that L'Hôpital's Rule cannot be used near \(a\)?
- In the Cauchy Mean Value Theorem argument, why must \(g'(c_x)\) be nonzero?
- Does failure of the standard hypotheses prove that the original quotient has no limit? Explain briefly.
