Find $\lim_{x\to 0^{+}} \frac{\theta(x)}{x}$

Question

Suppose $f\in C^1([0,1])$ and $f'(0)\neq 0$. For $x\in(0,1]$, let $\theta(x)$ be such that $$\int_0^x f(t)dt = f(\theta(x))x$$ Find $$\lim_{x\to 0^{+}} \frac{\theta(x)}{x}$$

I thinking about Taylor series expansion of $$F(x)=\int_0^x f(x)dx$$ but the point is the remainder term. I can't figure out which remainder term should I use. And what is the point of taking only $x\to 0^+$. Any kind of help is appreciable.

Answer 1

Note that $f$ is continuously differentiable in $[0,1]$ and $f'(0)\neq 0$ so that $f'$ maintains a constant sign in $[0,h] $ and hence $f$ is invertible in $[0,h]$ with inverse $g$ (say).

Next let $$F(x) =\int_{0}^{x}f(t)\,dt$$ and by definition we have $$\theta(x) =g\left (\frac{F(x)} {x} \right) $$ By Fundamental Theorem of Calculus we have $F(x) /x\to f(0)$ as $x\to 0^+$ and by continuity of $g$ this means that $\theta(x) \to g(f(0))=0$. Defining $\theta(0)=0$ our job is now to find $\theta'(0)$.

We have $$\theta'(x) =g'\left(\frac{F(x)} {x} \right) \cdot\frac{xf(x) - F(x)} {x^2}$$ Taking limits as $x\to 0^{+}$ and noting that $\theta, g$ are continuously differentiable we have $$\theta'(0)=g'(f(0))\cdot\lim_{x\to 0^{+}}\frac{xf(x)-F(x)}{x^2}=\frac{1}{f'(0)}\cdot\lim_{x\to 0^+}\frac{f(x)+xf'(x)-f(x)}{2x}=\frac{1}{2}$$ (last step uses L'Hospital's Rule).

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This question reminds me of the famous result in differential calculus which deals with limiting behavior of parameter $\theta$ which appears in Taylor's Theorem.

Let's first state it as

Theorem: Let $f$ be a real valued function defined in some neighborhood of $a$ such that $f^{(n+1)}$ is continuous in that neighborhood and $f^{(n+1)}(a)\neq 0$. If $$f(a+h) =f(a) +hf'(a) +\dots+\frac{h^{n-1}}{(n-1)!}f^{(n-1)}(a)+\frac{h^n}{n!}f^{(n)} (a+\theta h) $$ then $\theta\to 1/(n+1)$ as $h\to 0$.

And now to the proof of the above result. By Taylor's theorem we have $$f(a+h) =f(a) +hf'(a) +\dots+\frac{h^{n-1}}{(n-1)!}f^{(n-1)}(a)+\frac{h^n}{n!}f^{(n)} (a+\theta_n h) \tag{1}$$ and $$f(a+h) =f(a) +hf'(a) +\dots+\frac{h^{n-1}}{(n-1)!}f^{(n-1)}(a)+\frac{h^n}{n!}f^{(n)} (a)+\frac{h^{n+1}}{(n+1)!}f^{(n+1)}(a+\theta_{n+1}h) \tag{2}$$ where both $\theta_n, \theta_{n+1}$ lie in $(0,1)$. The subscript notation is used to distinguish the thetas appearing in the Taylor expansions above and the theorem mentioned above deals with $\theta_n$.

Comparing the two Taylor expansions above we get $$f^{(n)} (a+\theta_n h) =f^{(n)} (a) +\frac{hf^{(n+1)}(a+\theta_{n+1}h)}{n+1}\tag{3}$$ But using mean value theorem we have $$f^{(n)} (a+\theta_n h) =f^{(n)} (a) +\theta_n hf^{(n+1)}(a+\theta\theta_n h) \tag{4}$$ for some $\theta\in(0,1)$.

Again comparing $(3)$ and $(4)$ we get $$\theta_n=\frac{f^{(n+1)}(a+\theta_{n+1}h)}{(n+1) f^{(n+1)}(a+\theta\theta_{n+1}h)}$$ Letting $h\to 0$ we get $$\theta_n\to\frac{f^{(n+1)}(a)}{(n+1)f^{(n+1)}(a)}=\frac{1}{n+1}$$

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For your question apply the theorem to the anti-derivative $F$ with $a=0,n=1$ and use symbol $x$ in place of $h$. We have $$F(x) =F(0)+xF'(\theta_1 x) $$ ie $$\int_{0}^{x}f(t)\,dt=xf(\theta_1 x) $$ so that $\theta_1=\theta(x) /x$ and by the theorem this tends to $1/(n+1)=1/2$ as $x\to 0$ (provided $F''$ is continuous in neighborhood of $0$ and $F''(0)=f'(0)\neq 0$).

Answer 2

Applying L'Hopital's rule, we have

$$\lim_{x \to 0+}\frac{F(x) - f(0)x}{x^2} = \lim_{x \to 0+}\frac{F'(x) - f(0)}{2x} = \lim_{x \to 0+}\frac{f(x) - f(0)}{2x}= \frac{f'(0)}{2}$$

We also have

$$\frac{F(x) - f(0)x}{x^2} = \frac{f(\theta(x))- f(0)}{x} = \frac{f(\theta(x))- f(0)}{\theta(x)}\cdot \frac{\theta(x)}{x}$$

Thus,

$$\tag{*}\lim_{x \to 0+}\frac{f(\theta(x))- f(0)}{\theta(x)}\cdot \frac{\theta(x)}{x} = \frac{f'(0)}{2} $$

Since $\theta(x) \in (0,x)$, we must have $\theta(x) \to 0$ as $x \to 0+$ and

$$\lim_{x \to 0+}\frac{f(\theta(x))- f(0)}{\theta(x)} = f'(0)$$

Along with (*) this shows that

$$\lim_{x \to 0+} \frac{\theta(x)}{x} = \frac{1}{2}$$

Answer 3

• As $f'(0)\neq0$, $f$ is strictly monotone in a neighborhood of $0$ in $[0,1]$, which contains an interval say $[0,\alpha]$, with $\alpha\leq 1$.
• By the inverse function theorem, $f$ is invertible and differentiable in a puncture neighborhood of $0$. Without loss of generality, say that this neighborhood is also contains $(0,\alpha]$.
• The mean value theorem shows that for $0<x<\alpha$, $0<\theta(x)<x$; hence $\theta(x)\xrightarrow{x\rightarrow0}0$.
• Another way to establish the previous statement is by noticing that $f(\theta(x))=\frac{1}{x}\int^x_0f(s)\,ds\xrightarrow{x\rightarrow0}f(0)$. Then we can define $\theta(0)=0$.

Setting $F(x)=\int^x_0f(s)\,ds$, we have that $$ \begin{align} \theta(x)&=f^{-1}\big(\tfrac{F(x)}{x}\big), \quad 0<x\leq\alpha\\ \theta(0)&=0 \end{align} $$ defines a continuous function that is differentiable in $(0,\alpha]$.

The problem is now reduced to showing that $\theta$ is differentiable at $x=0$ and to estimate $\theta'(0)$.

Here we can appeal to the mean value theorem disguised as L'Hospital rule:

$$\begin{align} \frac{\theta(x)}{x}&\sim \frac{\theta'(x)}{1}=\frac{1}{f'(\theta(x))}\frac{xf(x)-F(x)}{x^2}\\ \end{align}$$

The factor $\frac{1}{f'(\theta(x))}\xrightarrow{x\rightarrow0}\frac{1}{f'(0)}$. Now we appeal once more to L'Hospital rule (all conditions (except the clariboyant one are satisfied) to the other factor:

$$ \begin{align} \frac{xf(x)-F(x)}{x^2}\sim\frac{f(x)+xf'(x)-f(x)}{2x}\xrightarrow{x\rightarrow0}\frac{f'(0)}{2} \end{align} $$

• This shows that the clairvoyant condition of L'Hospital rule is also satisfied. Now we may put things together to obtain $$\lim_{x\rightarrow0+}\frac{\theta(x)}{x}=\frac{1}{2}$$