When a Zero-Over-Zero Limit Needs More Than Substitution
Direct substitution is a useful first step for a limit. If it gives a quotient whose numerator and denominator both approach zero, however, substitution alone does not determine the limit. The quotient might approach a finite number, grow without bound, or fail to have a limit. This situation is called the indeterminate form \(0/0\).
L'Hôpital's Rule provides a way to evaluate some such limits. It replaces a quotient by the quotient of the derivatives of its numerator and denominator. The rule has specific hypotheses: the \(0/0\) form is essential, and the derivative quotient must have a limit. Checking those conditions is part of using the method, not an optional formality.
Statement of L'Hôpital's Rule for a Finite Point
Here is a two-sided version of the rule. The same statement can be applied on just one side of the point when the functions are defined only on that side.
The condition \(g'(x)\ne 0\) ensures that \(g(x)\ne 0\) near \(c\), except at \(c\), so the original quotient is defined there. In the common situation where \(f\) and \(g\) are given on a punctured interval and both approach zero at \(c\), they can be extended continuously by setting \(f(c)=g(c)=0\). The hypotheses can then be checked using those extensions.
The rule does not say that every \(0/0\) limit can be evaluated by differentiating. In particular, the limit of \(\frac{f'(x)}{g'(x)}\) must exist in the stated sense. If it does not, this version of the theorem gives no conclusion.
Why the Rule Works
The reason the derivative quotient controls the original quotient comes from the Cauchy Mean Value Theorem. The essential point is that the two quotients are equal at a nearby point \(\xi\), not necessarily at the same input \(x\).
Choose \(x\ne c\) close to \(c\). Since \(f(c)=g(c)=0\), the original quotient can be written as \(\frac{f(x)-f(c)}{g(x)-g(c)}\).
Continuity on the interval between \(c\) and \(x\), differentiability inside it, and \(g(x)\ne g(c)\) give a point \(\xi\) strictly between \(c\) and \(x\) such that \(\frac{f(x)-f(c)}{g(x)-g(c)}=\frac{f'(\xi)}{g'(\xi)}\).
As \(x\to c\), the point \(\xi\) between \(c\) and \(x\) also approaches \(c\). Thus, if \(\frac{f'(t)}{g'(t)}\to L\) as \(t\to c\), then \(\frac{f(x)}{g(x)}=\frac{f'(\xi)}{g'(\xi)}\to L\).
For completeness, \(g(x)\ne 0\) follows here from the hypotheses: if \(g(x)=g(c)=0\) for some nearby \(x\ne c\), the Mean Value Theorem applied to \(g\) on the interval between \(c\) and \(x\) would give \(g'(u)=0\) at an interior point \(u\), contradicting the assumption. This argument explains why the rule is a limit theorem, rather than an algebraic identity between quotients.
Worked Example: An Exponential Difference Quotient
Worked Example: An Exponential Difference Quotient
Evaluate \(\lim_{x\to 0}\frac{e^x-1}{x}\).
Step 1: Check the form. Substituting \(x=0\) gives \(e^0-1=0\) in the numerator and \(0\) in the denominator. The form is \(0/0\).
Step 2: Check the hypotheses. Let \(f(x)=e^x-1\) and \(g(x)=x\). Both functions are continuous at \(0\) and differentiable nearby, \(f(0)=g(0)=0\), and \(g'(x)=1\ne 0\).
Step 3: Differentiate separately. We have \(f'(x)=e^x\) and \(g'(x)=1\), so the derivative quotient is \(\frac{e^x}{1}=e^x\).
Step 4: Evaluate its limit. Since \(e^x\) is continuous, \(\lim_{x\to0}e^x=e^0=1\). The theorem therefore gives:
Check: The derivative quotient tends to \(1\), and every hypothesis required for this application has been verified. Thus the original limit is \(1\).
Worked Example: A Natural Logarithm Quotient
Worked Example: A Natural Logarithm Quotient
Evaluate \(\lim_{x\to0}\frac{\ln(1+x)}{x}\).
Step 1: Check the form and domain. At \(x=0\), the numerator is \(\ln(1)=0\), and the denominator is \(0\). The logarithm is defined for \(x>-1\), so the function is defined on a full interval around \(0\), apart from the quotient's excluded input \(x=0\).
Step 2: Check the derivative condition. Set \(f(x)=\ln(1+x)\) and \(g(x)=x\). They are continuous at \(0\), differentiable nearby, and both equal \(0\) there. Their derivatives are \(f'(x)=\frac{1}{1+x}\) and \(g'(x)=1\); in particular, \(g'(x)\ne0\).
Step 3: Evaluate the derivative quotient.
The last equality follows from continuity of \(\frac{1}{1+x}\) at \(0\). L'Hôpital's Rule now gives \(\lim_{x\to0}\frac{\ln(1+x)}{x}=1\).
Check: The original quotient has the required \(0/0\) form, and the derivative quotient has limit \(1\). The rule's conditions are satisfied.
Worked Example: A Square-Root Difference Quotient
Worked Example: A Square-Root Difference Quotient
Evaluate \(\lim_{x\to0}\frac{\sqrt{1+x}-1}{x}\).
Step 1: Check the form and domain. Substitution gives \(\sqrt{1}-1=0\) and \(x=0\), so the form is \(0/0\). The square root is defined when \(x\ge -1\), which includes an interval around \(0\).
Step 2: Verify the derivatives and their conditions. Let \(f(x)=\sqrt{1+x}-1\) and \(g(x)=x\). Both are continuous at \(0\), differentiable on an interval around \(0\), and equal \(0\) at \(0\). For inputs close to \(0\), \(f'(x)=\frac{1}{2\sqrt{1+x}}\), while \(g'(x)=1\ne0\).
Step 3: Find the derivative quotient limit.
By L'Hôpital's Rule, \(\lim_{x\to0}\frac{\sqrt{1+x}-1}{x}=\frac12\).
Check: The derivative quotient tends to \(\frac12\), and the original numerator and denominator both tend to zero. The verified hypotheses justify the conclusion.
A Common Mistake: Treating the Rule as an Identity
L'Hôpital's Rule relates limits; it does not state that \(\frac{f(x)}{g(x)}=\frac{f'(x)}{g'(x)}\) for each \(x\). For a concrete example, take \(f(x)=x^2\) and \(g(x)=x\), with \(x\ne0\). At \(x=\frac12\), the original quotient is \(\frac{x^2}{x}=\frac12\), but the derivative quotient is \(\frac{2x}{1}=1\). They are not equal at that input.
The correct conclusion concerns their limits as \(x\to0\). The original quotient is \(\frac{x^2}{x}=x\), so its limit is \(0\). The derivative quotient is \(2x\), whose limit is also \(0\). The Cauchy Mean Value Theorem proof explains the relationship: for each nearby \(x\), the original quotient matches the derivative quotient at some intermediate point \(\xi\), which need not equal \(x\).
Check Your Understanding
For each question, identify the relevant conditions before deciding whether L'Hôpital's Rule gives a conclusion.
- For \(\lim_{x\to0}\frac{\sin x}{x}\), what form results from substitution, and what derivative quotient should be evaluated?
- State the condition on \(g'(x)\) in the finite-point version of the rule given here.
- Why does the limit of \(\frac{f'(x)}{g'(x)}\), rather than its value at a single point, determine the conclusion?
- If the original quotient approaches \(0/0\) but the derivative quotient has no limit, what can you conclude from this version of L'Hôpital's Rule?
- For \(\lim_{x\to0}\frac{x^3}{x^2}\), evaluate both the original quotient's limit and the derivative quotient's limit. Are the quotients equal for every nonzero \(x\)?
