Let $f\in C^1(\mathbb{R})$ and $f(0)=0$. Show that $\frac{f(x)}{x}$ is in $C(\mathbb{R})$.

Question

I'm trying to solve the below problem. It's from an exam in real analysis. Thus, only such methods may be used.

Let $f\in C^1(\mathbb{R})$ and $f(0)=0$. Show that $\frac{f(x)}{x}$ is in $C(\mathbb{R})$.

This is my attempt. Is it correct?

We start by introducing the denotation $$g(x)=\frac{f(x)}{x}.$$ We may use L'Hôpital's rule since $f(0)=0$ and $x=0$, and both $f(x)$ and $x$ are continuous functions that form the quotient $0/0$ when $x=0$. Here \begin{align*} \lim_{x\to0}g(x)=\lim_{x\to0}\frac{f(x)}{x}=\lim_{x\to0}\frac{f'(x)}{1}=f'(x), \end{align*} and $f'(x)\in C(\mathbb{R})$ since $f(x)\in C^1(\mathbb{R})$.

The rule is applicable here as the numerator and the denominator only need to be differentiable separately, and the denominator needs to have a nonzero derivative on the domain. This also means that the limit only needs to be valid for the denominator and the numerator separately.

This is true here as the denominator is $x$, and the numerator belongs to $C^1(\mathbb{R})$ by assumption. Thus, the limit exists and is the derivative of the function at that point.

One way to do this is to say that by theorem a pair of continuous functions $f$ and $g$ form continuous functions $\alpha f+\beta g$, $fg$, or $f/g$, given that $g(x)\neq0$ for $f/g$, which was just covered.

Is there any other way to do the last part? I don't know if my professor will be fine with me just referring to some theorem in the book.

Answer 1

By the fundamental theorem and the chain rule, $$ f(x)=\int_0^1 f'(tx)x\,dt $$ so that $$ \frac{f(x)}{x}=\int_0^1 f'(tx)\,dt $$ and the right side is a continuous function of $x$.

Answer 2

The argument is almost correct. There's a typo in the line $$ \lim_{x\to0}g(x)=\lim_{x\to0}\frac{f(x)}{x}=\lim_{x\to0}\frac{f'(x)}{1}=f'(x), $$ it should be $$\lim_{x\to0}g(x)=\lim_{x\to0}\frac{f(x)}{x}=\lim_{x\to0}\frac{f'(x)}{1}=f'(0). $$ Then we are done by saying that $f'(0)=\frac{f(x)}{x}$, as Kavi Rama Murthy pointed out.

Another approach given by Saucy O'Path is to note that $$f'(0)=\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0}\frac{f(x)}{x},$$ and to define $$g(x)=\begin{cases} \frac{f(x)}{x} & \text{if}\ x\neq0 \\ f'(0) & \text{if}\ x=0 \end{cases}.$$ This is continuous as $f\in C^1(\Bbb{R})$ means that $f$ has a continuous derivative.

I hope that you are fine with me writing your comments as an answer to the question. If not, then I will delete this answer.

Answer 3

I suggest that the statement itself does not hold at all. Here is a counterexample.

Consider the situation when $f(x)=\sin x$. Then $f'(x)=\cos x \in C(\mathbb R),$ and $f(0)=0.$ But $\dfrac{f(x)}{x}=\dfrac{\sin x}{x}$ is not continuous in $\mathbb R$, because at least it has no definiton at $x=0$, even if $\lim\limits_{x \to 0}\dfrac{f(x)}{x}=1.$ Actually, the discontinuous point could be removed by making a supplementary definition, but this is not an inevitable conclusion could be drawn from the assumptions.

Now,I wonder how you so many people could prove a false statement.