Darboux, L'Hôpital, and Taylor's Theorem

For $n=0$, the displayed sum is $\iota \left(\genfrac{}{}{0px}{}{0}{0}\right)f^{(0)}{g}^{(0)}=fg=(fg{)}^{(0)}$.

Assume the formula holds at an index $k$, and assume $f,g$ have derivatives through order $k+1$.

Differentiating the finite sum gives $(fg{)}^{(k+1)}={\sum }_{j=0}^k\iota \left(\genfrac{}{}{0px}{}{k}{j}\right)(f^{(j+1)}{g}^{(k-j)}+f^{(j)}{g}^{(k-j+1)})$.

Shift $j+1$ in the first sum, retain $j$ in the second, and combine the two interior coefficients by Pascal's rule; the two boundary coefficients are $1$. The result is $(fg{)}^{(k+1)}={\sum }_{j=0}^{k+1}\iota \left(\genfrac{}{}{0px}{}{k+1}{j}\right)f^{(j)}{g}^{(k+1-j)}$.

Thus the formula holds for every $n\in \mathbb{N}$.

For $n=1$, Rolle's theorem on $[x_0,x_1]$ gives $c\in (x_0,x_1)$ with $f^{\mathrm{\prime }}(c)=0$.

For $n\ge 2$, apply Rolle on each $[x_{j-1},x_j]$ to obtain $y_j\in (x_{j-1},x_j)$ with $f^{\mathrm{\prime }}(y_j)=0$, so $y_1<\cdots <y_n$.

The function $f^{\mathrm{\prime }}$ is continuous on $[y_1,y_n]$, because those points lie in $(x_0,x_n)$ and the existence of $f^{\mathrm{\prime }\mathrm{\prime }}$ gives continuity there; it is $(n-1)$-times differentiable on $(y_1,y_n)$. Apply the induction hypothesis of order $n-1$ to $f^{\mathrm{\prime }}$ and the $n$ ordered zeros $y_1,\dots ,y_n$. This gives $c\in (y_1,y_n)\subset (x_0,x_n)$ with $(f^{\mathrm{\prime }}{)}^{(n-1)}(c)=f^{(n)}(c)=0$.

The claim follows for every $n\ge 1$.

If $\lambda =f^{\mathrm{\prime }}(x)$ or $\lambda =f^{\mathrm{\prime }}(y)$, choose that endpoint.

Suppose $f^{\mathrm{\prime }}(x)<\lambda <f^{\mathrm{\prime }}(y)$, and define $h(t)=f(t)-\lambda t$ on $[x,y]$. Then $h^{\mathrm{\prime }}(x)<0<h^{\mathrm{\prime }}(y)$.

If instead $f^{\mathrm{\prime }}(y)<\lambda <f^{\mathrm{\prime }}(x)$, apply the preceding argument to $-h$, obtaining an interior extremum of $h$.

For sufficiently small positive $s,t$, the derivative inequalities give $h(x+s)<h(x)$ and $h(y-t)<h(y)$. Hence a minimum of $h$ on $[x,y]$ occurs at an interior point $c$.

In either strict-order case, Fermat gives $h^{\mathrm{\prime }}(c)=0$, hence $f^{\mathrm{\prime }}(c)=\lambda$. Together with the endpoint case, every intermediate value is attained.

Assume $c$ is removable. Choose a value strictly between the common punctured limit and $g(c)$. On a sufficiently small punctured neighbourhood all values of $g$ lie on the limit side of that value, while the endpoint value lies on the other side, contradicting the intermediate value property on a segment ending at $c$.

Assume $c$ is a jump. The open interval between the unequal one-sided limits contains a value different from $g(c)$; choose such a value. Sufficiently close points on the two sides have values on opposite sides of the chosen value, while neither punctured side nor the point $c$ takes it, again contradicting the intermediate value property.

Thus neither kind of discontinuity can occur.

For any $a<b<c$ in $I$, $f(b)$ must lie strictly between $f(a)$ and $f(c)$. Indeed, if it lies above both, a value strictly between $\max \{f(a),f(c)\}$ and $f(b)$ is attained once in $(a,b)$ and once in $(b,c)$, contradicting injectivity; the case below both is analogous.

Fix $a<b$. If $f(a)<f(b)$, step 1.1 forces $f(x)<f(y)$ for every $x<y$ in $I$; inserting points between or beyond $a,b$ proves all possible placements. If $f(b)<f(a)$, the symmetric argument gives strict decrease.

Injectivity excludes equality, so one of the two alternatives holds and $f$ is strictly monotone.

If $f^{\mathrm{\prime }}$ is injective, [L1] and [L2] make it strictly monotone. If it is increasing, [L1] and [L3] make it continuous; if it is decreasing, apply [L3] to $-f^{\mathrm{\prime }}$, whose interval images are the negatives of the interval images of $f^{\mathrm{\prime }}$.

If $f^{\mathrm{\prime }}$ is nondecreasing, [L1] and [L3] make it continuous. If it is nonincreasing, the same argument applied to $-f^{\mathrm{\prime }}$ gives continuity.

These are the two stated alternatives.

If $g(a)=g(b)$, Rolle gives $d\in (a,b)$ with $g^{\mathrm{\prime }}(d)=0$, contrary to the hypothesis. Hence $g(b)-g(a)\mathrm{\ne }0$.

Cauchy's theorem supplies $c\in (a,b)$ with the cross-product identity in [L1].

Divide that identity by the two nonzero factors $g(b)-g(a)$ and $g^{\mathrm{\prime }}(c)$ to obtain the quotient formula.

Extend $f,g$ to $c$ by $f(c)=g(c)=0$. Their continuity at $c$ follows from the assumed zero limits, while differentiability gives continuity at every other point of the segment. For $x\mathrm{\ne }c$ sufficiently close, the quotient lemma on the segment with endpoints $c,x$ gives $\frac{f(x)}{g(x)}=\frac{f^{\mathrm{\prime }}({\xi }_x)}{g^{\mathrm{\prime }}({\xi }_x)}$, where ${\xi }_x$ lies strictly between $c$ and $x$.

As $x\to c$ in the chosen mode, ${\xi }_x\to c$ in that mode. Applying the defining finite or infinite limit inequality to the derivative quotient therefore gives $f(x)\mathrm{/}g(x)\to L$.

At infinity, put $F(t)=f(1\mathrm{/}t)$, $G(t)=g(1\mathrm{/}t)$. Then $F^{\mathrm{\prime }}\mathrm{/}{G}^{\mathrm{\prime }}=f^{\mathrm{\prime }}(1\mathrm{/}t)\mathrm{/}{g}^{\mathrm{\prime }}(1\mathrm{/}t)$, since the common factor $-1\mathrm{/}{t}^2$ cancels. Apply steps 1.1 and 2.1 as $t\to 0^{+}$ or $0^{-}$, and translate back.

Fix a base point $a$ inside the domain. For variable $x$ farther toward the limiting end, [L1] gives $\frac{f(x)-f(a)}{g(x)-g(a)}=\frac{f^{\mathrm{\prime }}({\xi }_x)}{g^{\mathrm{\prime }}({\xi }_x)}$, where ${\xi }_x$ lies between $a$ and $x$.

First choose $a$ sufficiently far toward the end that the derivative quotient is as close to $L$ as required throughout the remaining tail. Then the quotient of increments has the same bound for every later $x$.

Since $\mathrm{\mid }g(x)\mathrm{\mid }\to \mathrm{\infty }$, $g(a)\mathrm{/}g(x)\to 0$; since the increment quotient is bounded in the finite-$L$ case, the identity $\frac{f(x)}{g(x)}=\frac{f(x)-f(a)}{g(x)-g(a)}(1-\frac{g(a)}{g(x)})+\frac{f(a)}{g(x)}$ gives the finite conclusion. For $L=\pm \mathrm{\infty }$, choose the derivative-quotient lower or upper bound first and then make the two fixed-base terms negligible, obtaining the defining arbitrary bound.

Thus the quotient has limit $L$ in every stated mode.

At $r=0$ the formula is the definition.

Assuming the formula at $r<n$, differentiate termwise. The term indexed $j$ acquires $\iota (j-r)$, which cancels $\iota ((j-r)!)$ to $\iota ((j-r-1)!)$; the $j=r$ constant term disappears. This is the formula at $r+1$.

At $x=a$, only the $j=r$ term survives and equals $f^{(r)}(a)$. Subtraction from $f^{(r)}(a)$ gives the remainder assertion.

The formula and both consequences hold through order $n$.

Define $\mathrm{\Phi }(t):=f(x)-{\sum }_{j=0}^n{f}^{(j)}(t)(x-t{)}^j\mathrm{/}\iota (j!)$ and $\mathrm{\Psi }(t):=(x-t{)}^p$. Telescoping after differentiating the sum gives ${\mathrm{\Phi }}^{\mathrm{\prime }}(t)=-f^{(n+1)}(t)(x-t{)}^n\mathrm{/}\iota (n!)$, while ${\mathrm{\Psi }}^{\mathrm{\prime }}(t)=-\iota (p)(x-t{)}^{p-1}$.

We have $\mathrm{\Phi }(a)=R_{n,a}f(x)$, $\mathrm{\Phi }(x)=0$, $\mathrm{\Psi }(a)=(x-a{)}^p$, and $\mathrm{\Psi }(x)=0$. Also ${\mathrm{\Psi }}^{\mathrm{\prime }}\mathrm{\ne }0$ on $(a,x)$, because $p\ge 1$, $\iota (p)>0$, and $x-t>0$.

Apply [L2] to $\mathrm{\Phi },\mathrm{\Psi }$. For some $\xi \in (a,x)$, $\mathrm{\Phi }(a)\mathrm{/}\mathrm{\Psi }(a)={\mathrm{\Phi }}^{\mathrm{\prime }}(\xi )\mathrm{/}{\mathrm{\Psi }}^{\mathrm{\prime }}(\xi )=f^{(n+1)}(\xi )(x-\xi {)}^{n+1-p}\mathrm{/}(\iota (p)\iota (n!))$.

Multiply by $(x-a{)}^p$. If $x<a$, interchange the interval endpoints; the same algebraic identity results.

Set $p=n+1$ in [L1]. Then $(x-\xi {)}^0=1$ and $\iota (n+1)\iota (n!)=\iota ((n+1)!)$, giving the Lagrange form.

Set $p=1$. Since $\iota (1)=1$, the formula becomes the Cauchy form.

These are the asserted special cases.

Under the separate $n=0$ hypothesis, the quantified assertion is exactly the definition of continuity at $a$.

For $n=1$, the derivative definition gives $\frac{f(x)-f(a)-f^{\mathrm{\prime }}(a)(x-a)}{x-a}=\frac{f(x)-f(a)}{x-a}-f^{\mathrm{\prime }}(a)⟶0.$ This is the required base case.