Why One-Sided Boundary Limits Need Care
A function may be defined only to the right or only to the left of a point. For example, \(\sqrt{x}\) is real-valued for \(x\ge 0\), so a limit as \(x\to 0\) must be approached from the right. Likewise, \(\ln x\) is defined only for \(x>0\), and its behavior near \(0\) is described by a right-hand limit.
L’Hôpital’s Rule can apply in these settings, but its hypotheses must hold on the side from which \(x\) approaches the boundary. The rule does not require the functions to be defined on both sides, or even to be defined at the boundary itself. It does require an indeterminate form and appropriate differentiability and derivative conditions throughout a one-sided interval.
One-Sided Notation and the Required Conditions
The notation \(x\to a^+\) means that \(x\) approaches \(a\) through values greater than \(a\). The notation \(x\to a^-\) means that \(x\) approaches \(a\) through values less than \(a\). The point \(a\) itself need not belong to the domain.
The two forms in the theorem are called indeterminate because the information from substitution alone does not determine the quotient’s limit. In the zero-over-zero case, both parts tend to zero, but the relative rates are not yet known. In the infinity-over-infinity case, both magnitudes grow without bound, and again their relative rates matter.
Differentiability is required on the punctured interval, not necessarily at \(a\). The denominator must stay nonzero so that the quotient is defined. The derivative quotient must have a limit on the side in question; merely being able to compute \(f'(x)\) and \(g'(x)\) is not enough.
Why the Zero-Over-Zero Case Works
Here is a complete justification for the zero-over-zero case when \(x\to a^+\). Suppose \(f(x)\to 0\) and \(g(x)\to 0\) as \(x\to a^+\), and that \(f\) and \(g\) are differentiable to the right of \(a\), with \(g'(x)\ne 0\) there. Extend both functions to the endpoint by setting \(f(a)=g(a)=0\). These extensions are continuous at \(a\), because the one-sided limits are zero.
For any \(x>a\) sufficiently close to \(a\), the Cauchy Mean Value Theorem applied to \(f\) and \(g\) on \([a,x]\) gives a point \(c\) with \(a<c<x\) such that
Because \(f(a)=g(a)=0\), this becomes
As \(x\to a^+\), the point \(c\), which lies between \(a\) and \(x\), also approaches \(a\) from the right. Therefore, if \(f'(t)/g'(t)\to L\) as \(t\to a^+\), then \(f'(c)/g'(c)\to L\), and hence \(f(x)/g(x)\to L\). For \(x\to a^-\), apply the same argument on \([x,a]\). The interval is reversed in position, but the quotient of endpoint differences still simplifies to \(f(x)/g(x)\).
This argument shows why an endpoint value is not essential: in the zero-over-zero case, it can be supplied by continuous extension. It also shows why the derivatives need only be controlled on the side actually being approached.
Worked Example: A Right-Hand Limit at Zero
Worked Example: A Right-Hand Limit at Zero
Evaluate \(\displaystyle \lim_{x\to 0^+}\frac{e^x-1}{x}\).
Both functions are defined for \(x>0\) near zero. Substitution gives \(e^0-1=0\) in the numerator and \(0\) in the denominator, so the form is \(0/0\). The functions are differentiable on a right-hand punctured interval, and the denominator derivative is \(1\), which is nonzero. Apply L’Hôpital’s Rule:
Check: the defining difference quotient for the derivative of \(e^x\) at zero is \(\lim_{x\to0^+}(e^x-e^0)/(x-0)\). The value found is \(1\), agreeing with \( (e^x)'|_{x=0}=e^0=1\).
Worked Example: A Left-Hand Limit at a Boundary
Worked Example: A Left-Hand Limit at a Boundary
Evaluate \(\displaystyle \lim_{x\to 2^-}\frac{\ln(3-x)}{x-2}\).
For \(x\) near \(2\) from the left, \(3-x>0\), so the logarithm is defined. Substitution gives \(\ln(1)=0\) and \(2-2=0\), hence the form is \(0/0\). On a sufficiently small interval to the left of \(2\), both functions are differentiable, and the denominator derivative is \(1\). Differentiate numerator and denominator:
Thus
Check: set \(h=2-x\), so \(h\to0^+\), \(3-x=1+h\), and \(x-2=-h\). The original quotient becomes \(-\ln(1+h)/h\), which tends to \(-1\) because the derivative of \(\ln u\) at \(u=1\) is \(1\). This agrees with the result.
Worked Example: Both Magnitudes Diverge at a Boundary
Worked Example: Both Magnitudes Diverge at a Boundary
Evaluate \(\displaystyle \lim_{x\to 0^+}\frac{\ln x}{1/x}\).
The domain is \(x>0\). As \(x\to0^+\), \(\ln x\to-\infty\), while \(1/x\to+\infty\). Both magnitudes tend to infinity, so this is an infinity-over-infinity form. The functions are differentiable for \(x>0\), the denominator derivative \(-1/x^2\) is nonzero there, and the derivative quotient is
Therefore,
Check: the original quotient simplifies to \(x\ln x\). With \(t=1/x\), this is \(-\ln(t)/t\), which tends to zero as \(t\to\infty\) by the comparison of logarithmic and polynomial growth. The simplified expression confirms the limit.
Common Pitfall: Differentiating Without an Indeterminate Form
One-sidedness does not make L’Hôpital’s Rule applicable to every quotient near a boundary. The original quotient must first have one of the required forms. For example, consider \(\lim_{x\to0^+}\sqrt{x}/(1+x)\). Substitution gives \(0/1\), not \(0/0\) or infinity-over-infinity. The limit is already determined as zero by continuity of the numerator and denominator.
The correction is to evaluate the original expression’s form before differentiating:
The derivative quotient answers a useful limit question only when the hypotheses of L’Hôpital’s Rule are satisfied. It is not a replacement for the original quotient in every limit problem.
A Reliable One-Sided Checklist
At a domain boundary, write down the side of approach before applying the rule. Then check the relevant conditions in order:
Use \(x\to a^+\) or \(x\to a^-\), and identify a punctured interval on that side where the functions are defined.
Substitute the endpoint into the limiting behavior. Confirm \(0/0\) or that both magnitudes tend to infinity.
Verify differentiability on that one-sided interval, that \(g\) is nonzero for the quotient, and that \(g'\ne0\) nearby.
If its limit exists, L’Hôpital’s Rule gives the limit of the original quotient.
Before applying the rule again, determine the new form and verify the conditions again.
One-sided limits are especially important when a formula has a restricted domain. The same point can have different left- and right-hand behavior, or one of those limits may not be defined at all. L’Hôpital’s Rule respects that domain: it uses only the behavior, derivatives, and indeterminate form on the side where the approach is possible.
Check Your Understanding
For each question, state which side of the endpoint is relevant and identify the conditions that matter before applying L’Hôpital’s Rule.
- In \(\lim_{x\to0^+}\frac{\sin x}{x}\), what form results from substitution, and what is the derivative quotient?
- Why does the zero-over-zero proof extend the functions continuously to the endpoint?
- For \(\lim_{x\to1^-}\frac{\ln x}{x-1}\), which interval must be used to check differentiability?
- Can L’Hôpital’s Rule be applied directly to \(\lim_{x\to0^+}\frac{x}{1+x}\)? Explain using the form after substitution.
- In an infinity-over-infinity application, why is it not enough merely to know that \(f'\) and \(g'\) exist?
