Indeterminate forms and l’Hôpital’s rule
“May we not call them the ghosts of departed quantities?”
The form tells you what is unresolved
Substitution sometimes gives a value immediately. At other times, it tells us only that competing tendencies remain unresolved. The indeterminate form \(0/0\) is one such situation: both quantities vanish, but their relative rates of vanishing determine the quotient.
For instance, as \(x\to0^+\), \(x^2/x\to0\), \(x/x\to1\) and \(x/x^2\to+\infty\). The same formal shape has three different outcomes. Indeterminate forms like \(0/0\) and \(\infty/\infty\) are diagnoses, not answers.
A quotient whose numerator and denominator each tend to \(+\infty\) or \(-\infty\) has an infinity-over-infinity form; their signs may differ. Merely being unbounded is not enough: each must tend to an infinite limit. Again, the relative growth matters.
Theorem
L’Hôpital’s rule
Consider a one-sided limit as \(x\) approaches a finite endpoint, or a limit as \(x\to+\infty\) or \(x\to-\infty\). On the relevant open interval sufficiently near that endpoint, assume that \(f\) and \(g\) are differentiable and that \(g(x)\ne0\) and \(g^{\prime}(x)\ne0\).
All limits below use the same approach direction. Neither function needs to be defined at a finite approach point.
Suppose either both \(f\) and \(g\) tend to zero, or both tend to infinity in magnitude. If
\[\lim_{x\to a}\frac{f^{\prime}(x)}{g^{\prime}(x)}=L\]exists as a finite or infinite limit, then
\[\lim_{x\to a}\frac{f(x)}{g(x)}=L.\]The symbol \(a\) also represents the relevant infinite endpoint. For a two-sided finite limit, apply the one-sided statement on both sides and check that the answers agree.
The full theorem is stated without proof. Its general proof needs an extension of the mean value theorem; the explanation below proves only a restricted case.
Why derivatives enter a zero-over-zero limit (optional)
Suppose \(f(a)=g(a)=0\), both functions are differentiable at the finite input \(a\), and \(g^{\prime}(a)\ne0\). For \(x\ne a\),
\[\frac{f(x)}{g(x)}=\frac{[f(x)-f(a)]/(x-a)}{[g(x)-g(a)]/(x-a)}.\]Taking the derivative limits gives \(f^{\prime}(a)/g^{\prime}(a)\). If additionally the derivatives are continuous at \(a\), the derivative ratio has this same limit. This motivates the theorem in a simple setting, but does not prove cases where both derivatives vanish, derivatives fail at the endpoint, or the endpoint is infinite.
Use the rule as a conditional step
First compute the numerator and denominator limits separately. If the quotient has an eligible form and the hypotheses hold, differentiate numerator and denominator separately. Then evaluate the new ratio. Reapply only after checking the new form and hypotheses.
Example 1
A second application
Evaluate \(\displaystyle\lim_{x\to0}\frac{e^x-1-x}{x^2}\). Both parts tend to zero, and the functions are differentiable near zero. The denominator derivative \(2x\) is nonzero off zero. Thus we may study
\[\lim_{x\to0}\frac{e^x-1}{2x}.\]This is again \(0/0\); its denominator derivative is now the nonzero constant \(2\). A second application gives
\[\begin{aligned} &\lim_{x\to0}\frac{e^x-1-x}{x^2}\\ &\quad=\lim_{x\to0}\frac{e^x-1}{2x}\\ &\quad=\lim_{x\to0}\frac{e^x}{2}=\frac12. \end{aligned}\]The reasoning runs back from the last, existing limit to justify each earlier equality.
Example 2
Unbounded growth at different rates
As \(x\to+\infty\), both \(\ln x\) and \(x\) tend to infinity. Since the denominator derivative is \(1\),
\[\lim_{x\to+\infty}\frac{\ln x}{x}=\lim_{x\to+\infty}\frac{1/x}{1}=0.\]The logarithm remains unbounded; the zero limit says its size becomes small relative to \(x\).
Products and differences must be rewritten
Example 3
A vanishing factor times an unbounded factor
For \(x\to0^+\), the expression \(x\ln x\) has form \(0\cdot(-\infty)\). The rule cannot be applied directly to a product. Rewrite it as an unbounded quotient:
\[\begin{aligned} \lim_{x\to0^+}x\ln x &=\lim_{x\to0^+}\frac{\ln x}{1/x}\\ &=\lim_{x\to0^+}\frac{1/x}{-1/x^2}\\ &=\lim_{x\to0^+}(-x)=0. \end{aligned}\]Here the derivative of \(1/x\) is nonzero on \(x>0\), so the quotient meets the conditions.
Example 4
A difference of two unbounded terms
Evaluate \(\displaystyle\lim_{x\to0^+}\left(\frac1x-\frac1{e^x-1}\right)\). Both terms tend to positive infinity. Combine them:
\[\frac1x-\frac1{e^x-1}=\frac{e^x-1-x}{x(e^x-1)}.\]This has form \(0/0\). The first derivative ratio is
\[\frac{e^x-1}{e^x-1+xe^x},\]again \(0/0\). The second derivative ratio is \(e^x/(2e^x+xe^x)\), which tends to \(1/2\). Each denominator derivative is positive for \(x>0\), so the applications are valid. The original difference tends to \(1/2\).
Powers: take a logarithm first
For a positive base \(u(x)\), put \(y=u(x)^{v(x)}\). The logarithm turns a changing exponent into a product:
\[\ln y=v(x)\ln(u(x)).\]The forms \(1^\infty\), \(0^0\) and \(\infty^0\) may become indeterminate products or quotients. After finding the logarithm limit, return to \(y\) using continuity of the exponential.
Example 5
A base approaching one
Find \(\displaystyle\lim_{x\to0}(1+2x)^{1/x}\). Near zero the base is positive. For \(y=(1+2x)^{1/x}\),
\[\begin{aligned} \lim_{x\to0}\ln y &=\lim_{x\to0}\frac{\ln(1+2x)}x\\ &=\lim_{x\to0}\frac{2}{1+2x}=2. \end{aligned}\]Therefore \(\displaystyle\lim_{x\to0}(1+2x)^{1/x}=e^2\). The number \(2\) is the limit of the logarithm, not of the original power.
Warning
A nonzero numerator over a vanishing denominator is not an eligible indeterminate form. For example, differentiating numerator and denominator in \(1/x\) would incorrectly suggest zero.
The rule concerns limits, not equality of the original functions with their derivative ratio. If the derivative ratio has no limit, this version of the theorem gives no conclusion; the original quotient may still converge. Algebra, squeeze arguments and standard limits remain useful.
A convergent quotient with an oscillating derivative ratio (optional)
As \(x\to+\infty\), let \(f(x)=x+\sin x\) and \(g(x)=x\). Both tend to \(+\infty\), but
\[\frac{f^{\prime}(x)}{g^{\prime}(x)}=1+\cos x\]has no limit. Nevertheless,
\[\frac{f(x)}{g(x)}=1+\frac{\sin x}{x}\longrightarrow1,\]because \(-1/x\le\sin x/x\le1/x\) for \(x>0\). The squeeze theorem settles the limit when l’Hôpital’s rule gives no conclusion.