Why the Side of Approach Matters
A function need not be defined on both sides of a point. For example, \(\sqrt{x}\) is real-valued only when \(x\ge 0\), so its behavior near \(0\) can be studied from the right but not from the left. L’Hôpital’s Rule still applies to eligible one-sided limits: its derivative conditions need to hold on the interval from which \(x\) approaches the point.
The key is to check the actual approach interval. For \(x\to a^+\), that is an interval \((a,a+\delta)\); for \(x\to a^-\), it is \((a-\delta,a)\), where \(\delta>0\). The functions and derivatives need not be defined on the other side of \(a\), or even at \(a\).
The One-Sided Rule
Suppose \(f\) and \(g\) are differentiable on a one-sided interval next to \(a\), and the quotient \(f(x)/g(x)\) is defined there. If the quotient has an indeterminate form \(0/0\) or \(\infty/\infty\) as \(x\) approaches \(a\) from that side, and \(g'(x)\ne 0\) throughout a sufficiently small approach interval, L’Hôpital’s Rule permits replacing the quotient by \(f'(x)/g'(x)\), provided the latter limit exists (finite or infinite).
For a \(0/0\) form, the one-sided limits of both \(f\) and \(g\) must be \(0\). For an \(\infty/\infty\) form, the absolute values of both functions must tend to infinity; their signs do not have to match. In either case, the derivative quotient must be considered on the same side of \(a\). If its one-sided limit is \(L\), then the original quotient has one-sided limit \(L\).
The hypotheses concern values arbitrarily close to the endpoint, not at the endpoint. In a \(0/0\) limit, \(f\) and \(g\) can be extended continuously to \(a\) by assigning them the value \(0\); the derivatives are still required only in the open approach interval.
Why the Rule Works for a One-Sided \(0/0\) Limit
Here is a justification for the right-sided case. The left-sided argument is the same, using points to the left of \(a\). Suppose \(\lim_{x\to a^+}f(x)=\lim_{x\to a^+}g(x)=0\). Extend both functions continuously to \(a\) by setting \(f(a)=g(a)=0\). For any \(x>a\) close enough to \(a\), the Cauchy Mean Value Theorem applies on \([a,x]\). It gives a point \(c\) with \(a<c<x\) such that
Since \(f(a)=g(a)=0\), the left side is \(f(x)/g(x)\). As \(x\to a^+\), the point \(c\), which lies between \(a\) and \(x\), also approaches \(a\) from the right. Therefore, if \(f'(x)/g'(x)\to L\) from the right, then \(f'(c)/g'(c)\to L\), and consequently \(f(x)/g(x)\to L\). This argument also works when the derivative quotient tends to \(+\infty\) or \(-\infty\).
The proof makes the domain requirement clear: every point used lies in \([a,x]\), on the right-hand side of the endpoint. No information about the functions for \(x<a\) is needed.
Worked Examples
Worked Example: A Square Root Quotient at the Domain Endpoint
Evaluate \(\lim_{x\to 0^+}\frac{\sqrt{x}}{x}\).
Step 1: Check the approach interval and form. Both \(\sqrt{x}\) and \(x\) are defined for \(x>0\). As \(x\to0^+\), both tend to \(0\), so the quotient has form \(0/0\).
Step 2: Check the derivatives on that interval. For \(x>0\), \(f'(x)=\frac{1}{2\sqrt{x}}\) and \(g'(x)=1\). These derivatives exist for every \(x>0\), and \(g'(x)\ne0\).
Step 3: Apply the rule from the right.
Check: For \(x>0\), \(\frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}}\), which tends to \(+\infty\) as \(x\to0^+\). This agrees with the derivative quotient.
Worked Example: Repeated Applications on a Right-Sided Interval
Evaluate \(\lim_{x\to0^+}\frac{1-\cos x}{x^2}\).
Step 1: Check the original form. As \(x\to0^+\), \(1-\cos x\to0\) and \(x^2\to0\). Both functions are differentiable for \(x>0\), and the denominator is nonzero there.
Step 2: Apply L’Hôpital’s Rule once. The derivatives are \(\sin x\) and \(2x\). The resulting quotient still has form \(0/0\), and \(2x\ne0\) for \(x>0\), so another application is permitted.
Step 3: Apply the rule again. The derivatives of \(\sin x\) and \(2x\) are \(\cos x\) and \(2\), respectively. The new denominator derivative is nonzero, so
Check: Using \(1-\cos x=2\sin^2(x/2)\), the original quotient is \(\frac{1}{2}\left(\frac{\sin(x/2)}{x/2}\right)^2\), which tends to \(\frac12\). All differentiations and limit evaluations used values with \(x>0\).
Worked Example: An Infinite-over-Infinite Form on a Restricted Domain
Evaluate \(\lim_{x\to0^+}\frac{\ln x}{1/x}\).
Step 1: Check the form and domain. Both functions are defined for \(x>0\). As \(x\to0^+\), \(\ln x\to-\infty\) and \(1/x\to+\infty\), so the quotient has an infinite-over-infinite form.
Step 2: Differentiate on the right-sided interval. For every \(x>0\), \((\ln x)'=1/x\) and \((1/x)'=-1/x^2\). The denominator derivative is nonzero there.
Step 3: Apply L’Hôpital’s Rule.
Check: The quotient simplifies to \(x\ln x\), which tends to \(0\) as \(x\to0^+\). For example, setting \(t=1/x\) gives \(x\ln x=-\ln(t)/t\), which tends to \(0\) as \(t\to+\infty\). The one-sided derivative calculation gives the same result.
Worked Example: A Left-Sided Logarithmic Limit
Evaluate \(\lim_{x\to0^-}\frac{\ln(1-x)}{x}\).
Step 1: Check the left-sided domain and form. If \(x<0\), then \(1-x>1\), so \(\ln(1-x)\) is defined. As \(x\to0^-\), the numerator and denominator both tend to \(0\).
Step 2: Differentiate on the left. For \(x<0\) near \(0\), the derivatives are \(-1/(1-x)\) and \(1\). The denominator derivative is nonzero.
Check: The standard limit \(\lim_{u\to0}\ln(1+u)/u=1\), with \(u=-x\), gives \(\ln(1-x)/x=-\ln(1+u)/u\to-1\). The left-sided result is confirmed.
A Practical Hypothesis Check
Before using L’Hôpital’s Rule on a one-sided limit, check each condition on the side actually being approached. A compact checklist is:
Write down whether \(x\to a^+\) or \(x\to a^-\), and choose an interval on that side where the functions are defined.
Check that the numerator and denominator both tend to \(0\), or that both have absolute values tending to infinity.
Confirm that both functions are differentiable on the open approach interval and that \(g'(x)\ne0\) there.
If applying the rule again, repeat the form and hypothesis checks for the new quotient.
The endpoint itself is not part of the open interval on which derivatives are required. This matters particularly for square roots and logarithms, whose domains may have an endpoint at the limiting point. The derivative formula need only make sense at nearby points in the permitted domain.
Common Mistake: Requiring Derivatives at the Endpoint
A common error is to reject \(\lim_{x\to0^+}\sqrt{x}/x\) because the derivative \((\sqrt{x})'=1/(2\sqrt{x})\) is undefined at \(x=0\). That checks the wrong point. L’Hôpital’s Rule requires differentiability on an open interval of approach, such as \((0,\delta)\), not at \(0\).
The correction is to check all points with \(0<x<\delta\). There, \(\sqrt{x}\) is differentiable, its derivative is \(1/(2\sqrt{x})\), and the denominator function \(g(x)=x\) has derivative \(g'(x)=1\ne0\). The quotient is \(0/0\) in the right-sided limit, so the rule applies and gives \(+\infty\). No derivative at the endpoint is needed.
Why Careful Side-Checking Matters
A two-sided limit exists only if the left- and right-sided limits both exist and are equal. But a restricted-domain function may have only one relevant side; \(\sqrt{x}\) near \(0\) is a basic example. In that situation, a right-sided limit can be meaningful even though a left-sided limit is not defined in the real domain.
One-sided reasoning also prevents accidental use of a derivative formula where its input is outside the domain. For example, when \(x\to0^+\), derivatives involving \(\sqrt{x}\) or \(\ln x\) should be evaluated only for \(x>0\). The rule does not extend the domain; it compares the original quotient and its derivative quotient along the interval where the functions already exist.
Check Your Understanding
For each limit, identify the approach interval and decide whether the stated L’Hôpital step is justified before evaluating it.
- For \(\lim_{x\to0^+}\frac{\sqrt{x}}{\sqrt{x}+x}\), what is the form, and are both functions differentiable on a right-sided interval?
- For \(\lim_{x\to0^-}\frac{\ln(1+x)}{x}\), which side of \(0\) is in the domain, and what derivative quotient results?
- For \(\lim_{x\to0^+}\frac{x}{\ln x}\), does the original quotient have form \(0/0\) or \(\infty/\infty\)?
- Why does the failure of \(f'(a)\) to exist not by itself prevent applying the rule to \(x\to a^+\)?
- When applying L’Hôpital’s Rule a second time, which conditions must be checked again?
