L Hopital rules with complete tail arguments

Proof

Prerequisite proofs used below

Fix an integer $n\ge1$ and the Euclidean norm on $\mathbb R^n$. A set is bounded if it lies in some finite-radius ball, closed if its complement is open, and compact if every cover by open subsets of $\mathbb R^n$ has a finite subcover. We use real completeness, including the nested closed interval property.

Heine–Borel in finite-dimensional Euclidean space

Theorem. A set $K\subseteq\mathbb R^n$ is compact if and only if it is closed and bounded. The empty set satisfies all three properties.

Compact implies bounded. The open balls $B(0,m)$, $m\ge1$ an integer, cover $K$. A finite subcover puts $K$ inside the largest of these balls.

Compact implies closed. Suppose $K\ne\varnothing$ and $y\notin K$. The balls $B(x,\|x-y\|/2)$, $x\in K$, cover $K$. Choose a finite subcover with centers $x_1,\ldots,x_r$ and set $\delta=\min_i\|x_i-y\|/2>0$. If $z$ lies in the ball centered at $x_i$, the triangle inequality gives $\|z-y\|>\|x_i-y\|/2\ge\delta$. Hence $B(y,\delta)$ misses $K$, so the complement is open.

Closed and bounded implies compact. Let nonempty $K$ be closed and bounded. Enclose it in a closed cube $Q_0=[-M,M]^n$ with $M>0$. Suppose an open cover of $K$ has no finite subcover. Divide each coordinate interval into two closed halves. The resulting $2^n$ closed subcubes cover $Q_0$. At least one subcube has intersection with $K$ that is not finitely coverable by the given family; otherwise the finite union of the finite subcovers would cover $K$. Select such a subcube and repeat. This produces nested closed cubes $Q_j$ with side length $2M/2^j$ and $K\cap Q_j$ not finitely coverable, hence nonempty.

For each coordinate, the nested interval property gives a common coordinate value. Together these give $p\in\bigcap_jQ_j$. Every pair of points in $Q_j$ has distance at most $2M\sqrt n/2^j$, so the intersection contains only $p$. Choose $x_j\in K\cap Q_j$. Then $\|x_j-p\|\le2M\sqrt n/2^j\to0$. A closed set contains such a limit: if $p$ were in its open complement, a ball around $p$ would eventually contain $x_j$ outside $K$, a contradiction. Thus $p\in K$.

One member $O$ of the cover contains $p$, and openness gives $B(p,\varepsilon)\subseteq O$ for some $\varepsilon>0$. For sufficiently large $j$, every point of $Q_j$ is within $\varepsilon$ of $p$, so $O$ alone covers $K\cap Q_j$. This contradicts the selection of $Q_j$. Therefore every open cover has a finite subcover. The number $2^n$ is finite; replacing it by infinitely many subpieces would invalidate this argument.

Continuous images and extreme values

Theorem. If $K\subseteq\mathbb R^n$ is compact and $F:K\to\mathbb R^m$ is continuous relative to $K$, then $F(K)$ is compact. In particular, a continuous $f:K\to\mathbb R$ on nonempty compact $K$ is bounded and attains its maximum and minimum.

Proof. An open cover $(O_i)$ of $F(K)$ pulls back to relatively open sets $F^{-1}(O_i)=K\cap U_i$, with $U_i$ open in $\mathbb R^n$. These $U_i$ cover $K$, so finitely many suffice. Their corresponding $O_i$ cover $F(K)$, proving compactness. For scalar $f$, the image is nonempty, closed and bounded by the theorem already proved in dimension one. Let $a=\sup f(K)$. For each $j\ge1$, the definition of supremum supplies $y_j\in f(K)$ with $a-1/j<y_j\le a$. Then $y_j\to a$, and closedness places $a$ in the image. Thus the maximum is attained. Applying the same argument to $-f$ gives the minimum. This proves existence, not a formula for the maximizing input.

Fermat's necessary condition. At a differentiable interior local maximum or minimum, $f'(a)=0$.

Proof. At a local maximum, sufficiently small positive increments give nonpositive difference quotients and negative increments give nonnegative ones. A two-sided derivative can satisfy both sign limits only by being zero. Negating the function treats a minimum. $\square$

The interior hypothesis matters: $f(x)=x$ on $[0,1]$ has extrema at endpoints despite derivative 1. The converse fails: $x^3$ has zero derivative at zero but no extremum there.

Rolle's theorem. If $a<b$, $f$ is continuous on $[a,b]$, differentiable on $(a,b)$ and $f(a)=f(b)$, then $f'(c)=0$ for some $c\in(a,b)$.

Proof. The extreme-value theorem supplies maximum and minimum. If the function is constant, every interior point works. Otherwise a value differs from the equal endpoint values, so either a maximum or a minimum occurs in the interior. Fermat's condition applies there. $\square$

Theorem: Cauchy's mean value theorem

If $f,g$ are continuous on $[a,b]$ and differentiable on $(a,b)$ with $a<b$, some $c\in(a,b)$ satisfies

$$[f(b)-f(a)]g'(c)=[g(b)-g(a)]f'(c).$$

Proof. The auxiliary function $H(t)=[f(b)-f(a)]g(t)-[g(b)-g(a)]f(t)$ has equal endpoint values. Rolle's theorem gives $H'(c)=0$, precisely the identity. It is valid even when a denominator in a ratio formulation would be zero. Dividing to write derivative ratios requires separately checking those denominators; this is essential in proofs of L'Hopital's rule.

Learning goals and a precise one-sided setting

Identify genuine indeterminate forms, prove the rule rather than treating derivative division as algebra, and distinguish sufficient hypotheses from necessary ones. Let $f,g$ be differentiable on an open interval approaching an endpoint $b$, which may be finite or $+\infty$, from the left. Assume $g'$ is nonzero throughout a sufficiently late approach interval. The analogous right-hand approach and negative-infinity cases follow by reversing coordinates.

Cauchy's mean value theorem says that for distinct interior $x,y$, some $c$ between them satisfies

$$\frac{f(x)-f(y)}{g(x)-g(y)}=\frac{f'(c)}{g'(c)}.$$

The denominator $g(x)-g(y)$ cannot vanish, for otherwise Rolle's theorem on $g$ would give a zero derivative between them. This also shows $g$ takes any value, including zero, at most once there. By shortening the approach interval we may therefore ensure $g\ne0$, so the desired quotient is defined.

Theorem: finite derivative-ratio limit

If $f'/g'\to L\in\mathbb R$ as the endpoint is approached, then $f/g\to L$ provided either (i) both $f,g\to0$, or (ii) $|g|\to\infty$. In case (ii), no separate assumption that $f$ is unbounded is necessary.

Proof of (i). Let $h=f-Lg$. Given $\varepsilon>0$, throughout a sufficiently late interval $|h'/g'|<\varepsilon$. Cauchy's theorem gives $|h(x)-h(y)|\le\varepsilon|g(x)-g(y)|$ for any two points there. Fix $x$ and let $y$ approach the endpoint. Since $h(y),g(y)\to0$, this yields $|h(x)|\le\varepsilon|g(x)|$, so $|f(x)/g(x)-L|\le\varepsilon$. This argument works even when the endpoint is infinity, without pretending infinity is an endpoint value of a continuous function.

Proof of (ii). Given $\eta>0$, set $\varepsilon=\eta/4$. Again take $h=f-Lg$ and a late interval where the derivative ratio has magnitude below $\varepsilon$. Fix one point $x_0$ in it. Cauchy's theorem between $x_0$ and a later $x$ gives $|h(x)-h(x_0)|\le\varepsilon|g(x)-g(x_0)|$. Consequently

$$\left|\frac{f(x)}{g(x)}-L\right| \le\frac{|h(x_0)|}{|g(x)|}+\varepsilon\left|1-\frac{g(x_0)}{g(x)}\right|.$$

For sufficiently late $x$, the first term is less than $\eta/2$ and the absolute factor in the second is less than $2$. The total is therefore less than $\eta/2+2(\eta/4)=\eta$. Since $\eta>0$ was arbitrary, the quotient tends to $L$.

Theorem: infinite derivative-ratio limits

Under the same denominator assumptions, if $f'/g'\to+\infty$, then $f/g\to+\infty$ in either case (i) or (ii); replace signs for a limit of $-\infty$.

Proof. For any real threshold $M$, on a sufficiently late interval $f'/g'>M$. Cauchy's theorem makes every incremental ratio greater than $M$. In case (i), fix $x$ and pass the other point to the endpoint to obtain $f(x)/g(x)\ge M$. Since $M$ can be chosen above any requested threshold, this proves divergence to $+\infty$. In case (ii), fix $x_0$ and write the incremental ratio as $R(x)>M$:

$$\frac{f(x)}{g(x)}=R(x)\left(1-\frac{g(x_0)}{g(x)}\right)+\frac{f(x_0)}{g(x)}.$$

For $M>0$ and sufficiently late $x$, the multiplying factor exceeds $1/2$ and $f(x_0)/g(x)>-1$. Therefore $f(x)/g(x)>M/2-1$. Taking $M$ arbitrarily large proves divergence to $+\infty$. Applying the result to $-f$ covers the negative-infinity limit.

Worked examples and diagnosis

A power form needs a logarithm first. For $x\to\infty$, $x^{1/x}>0$ and its logarithm is $\ln x/x$. The denominator tends to infinity and the derivative quotient is $1/x\to0$, so the proven rule gives log limit zero. Exponential continuity gives $x^{1/x}\to1$. Directly differentiating the numerator and denominator of the original power expression would not be an application of the rule.

Not every quotient is indeterminate. As $x\to0$, $(1+x)/(2+x)\to1/2$ by continuity. Differentiating numerator and denominator would give one, an incorrect limit because neither zero-over-zero nor an unbounded denominator applies.

The rule is not reversible. For $x\to\infty$, let $f=x+\sin x$ and $g=x$. Then $f/g=1+\sin x/x\to1$, but $f'/g'=1+\cos x$ has no limit. A quotient may have a limit even when the derivative ratio does not.

Graded practice

1. Foundation. Evaluate $\lim_{x\to0}(e^x-1)/x$ with hypotheses stated.
2. Core. Evaluate $\lim_{x\to\infty}\ln x/\sqrt x$.
3. Core. Evaluate $\lim_{x\downarrow0}x\ln x$ by converting it to an admissible quotient.
4. Proof. Explain why infinitely many zeros of the denominator derivative approaching the endpoint, so that every tail interval contains one, prevent direct use of the stated theorem but do not by themselves prove the quotient has no limit. Give an example.

Full solutions

1. Both functions tend to zero; the denominator derivative is one and the derivative ratio is $e^x\to1$. Therefore the limit is one. This agrees with the derivative definition of the exponential at zero.
2. The denominator tends to infinity, its derivative is $1/(2\sqrt x)\ne0$, and the derivative quotient is $(1/x)/(1/(2\sqrt x))=2/\sqrt x\to0$. Hence the limit is zero.
3. Write $x\ln x=\ln x/(1/x)$. The denominator tends to infinity and its derivative is $-1/x^2\ne0$; the derivative quotient is $-x\to0$. The right-hand approach version of case (ii) gives limit zero.
4. The theorem requires a zero-free derivative on a whole approach interval, so repeated derivative zeros approaching the endpoint violate that sufficient hypothesis. Let $f(x)=g(x)=x+\sin x$ as $x\to\infty$. Then $g'=1+\cos x$ has infinitely many zeros, but $f/g$ is identically one for all sufficiently large $x$. Failure of a hypothesis means the theorem is unavailable, not that its desired conclusion is false.

Summary and term definitions

An indeterminate form names a limit pattern that does not decide a quotient by itself. A derivative-ratio rule is a sufficient limiting theorem, not cancellation of differentiation symbols. Eventual hypotheses must hold on an entire tail interval. Cauchy's theorem controls incremental ratios; endpoint limits or a growing denominator then remove the base values and justify the original quotient limit.