Darboux, L'Hôpital, and Taylor's Theorem: Examples and Counterexamples

On every interval $[k,k+1)$, $\psi$ is the same quartic in $u=t-k$, with derivative $32u(1-u)(1-2u)$. Its values and first derivatives at $u=0$ and $u=1$ are all $0$, so adjacent pieces and their derivatives agree continuously at every integer.

Translation by an integer leaves the fractional part unchanged, so $\psi$ is $1$-periodic. Step 1.1 and the polynomial formula prove $C^1$-regularity, and $0\le \psi \le 1$. At fractional parts $u=1\mathrm{/}4$ and $u=3\mathrm{/}4$, the derivative formula gives ${\psi }^{\mathrm{\prime }}=3$ and ${\psi }^{\mathrm{\prime }}=-3$, respectively.

Since $\psi$ is bounded, $F(x)\mathrm{/}x=x\psi (1\mathrm{/}x)\to 0$, so $F^{\mathrm{\prime }}(0)=0$ by The derivative $f^{\mathrm{\prime }}(c)={\lim }_{x\to c}\frac{f(x)-f(c)}{x-c}$ of $f:A\to \mathbb{R}$ at a point $c\in A$ that is a limit point of $A$, and differentiability on a set.

For $x\mathrm{\ne }0$, $F^{\mathrm{\prime }}(x)=2x\psi (1\mathrm{/}x)-{\psi }^{\mathrm{\prime }}(1\mathrm{/}x)$. The periodic piecewise-polynomial derivative ${\psi }^{\mathrm{\prime }}$ takes two separated values along sequences tending to infinity, so $F^{\mathrm{\prime }}(x)$ has no limit at $0$.

Thus $F$ is differentiable and $F^{\mathrm{\prime }}$ is discontinuous at $0$; [L2] also confirms that its oscillation is not a jump.

On $x>0$, $f_k=x^{k+1}$; on $x<0$, $f_k=-x^{k+1}$.

Differentiating $j\le k$ times gives constant multiples of $x^{k+1-j}$ with opposite signs, and both one-sided values tend to $0$. Defining the derivative value at $0$ by the difference quotient gives matching continuous derivatives through order $k$.

The $(k+1)$-st one-sided derivatives are $(k+1)!$ and $-(k+1)!$, so that derivative does not exist at $0$.

Hence $f_k\in C^k\setminus C^{k+1}$ according to Higher derivatives and the classes $C^k$ and $C^{\mathrm{\infty }}$.

Boundedness of $\psi$ gives $F^{\mathrm{\prime }}(0)={\lim }_{x\to 0}(1+x\psi (1\mathrm{/}x))=1>0$.

For $x\mathrm{\ne }0$, $F^{\mathrm{\prime }}(x)=1+2x\psi (1\mathrm{/}x)-{\psi }^{\mathrm{\prime }}(1\mathrm{/}x)$. Along reciprocal sequences at which ${\psi }^{\mathrm{\prime }}>2$ the derivative is eventually negative, while along reciprocal sequences at which ${\psi }^{\mathrm{\prime }}<-2$ it is eventually positive.

If $F$ were monotone on some neighbourhood, [L2] would force one weak derivative sign throughout it, contradicting step 1.2.

Numerator and denominator tend to $0$, the denominator derivative $2x$ is nonzero near $1$, and the derivative quotient $(3x^2-1)\mathrm{/}(2x)$ tends to $1$.

Applying [L1] gives the limit $1$. Direct factorization to $x$ away from $x=\pm 1$ confirms the removable nature of the quotient at $1$.

Both $f(x)$ and $g(x)$ tend to $0$, and $f(x)\mathrm{/}g(x)=x\psi (1\mathrm{/}x)\to 0$ because $\psi$ is bounded.

Yet $f^{\mathrm{\prime }}(x)\mathrm{/}{g}^{\mathrm{\prime }}(x)=2x\psi (1\mathrm{/}x)-{\psi }^{\mathrm{\prime }}(1\mathrm{/}x)$, which has no limit because ${\psi }^{\mathrm{\prime }}$ has separated recurring values.

The quotient limit exists while the derivative-quotient limit does not, so the converse fails.

Induction gives $f^{(j)}(x)=\iota (j!)(1-x{)}^{-j-1}$, hence $f^{(j)}(0)\mathrm{/}\iota (j!)=1$.

Multiplying ${\sum }_{j=0}^n{x}^j$ by $1-x$ telescopes to $1-x^{n+1}$. Subtracting from $1\mathrm{/}(1-x)$ gives the stated remainder.

This exact expression agrees with the qualitative estimate supplied by [L2] on every closed interval avoiding $1$.

Direct differentiation gives common first and second derivative data $0,0$ at the origin.

The fourth derivative is first nonzero for $\pm x^4$, with opposite signs; the third derivative is first nonzero for $x^3$.

The even and odd cases of [L2] yield the three stated behaviours, explicitly realizing [L1].