# Prove that $g(x)=xf(x)$ given $g$ is differentiable at $0$ and $g(0)=0$ and $f$ is some function continuous at $0$

May be, I am over thinking this problem. Over here, do we just construct a piece wise function $f$ such that is $g(x)/x$ for all $x \neq 0$ and it is equal to $0$ for $x=0$?

I was a little thrown off by the question. I just didn't see a rigorous enough way to do it. Any help will be appreciated.

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You need to set $f(x) = \frac{g(x)}{x}$ for $x\neq 0$ and define $f(0) := \lim_{x \to 0} g(x)/x$. To make this well-defined, show that this limit exists. This will imply that $f$ is continuous and satisfies the necessary properties.

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One question and please be charitable in answering this. Should the $f(0)$ be written as $f'(0)$? For it seems as if we use the limit def of derivatives, then we get the same expression. –  user43901 Jan 16 '13 at 4:21
@user43901: No, $f'(0)$ might not exist, and even if it does it is unlikely to equal $f(0)$. Remembering that $g(0)=0$, what is $\lim\limits_{x\to 0}\dfrac{g(x)}{x}$? In particular, how do we know that limit exists from what was given in the problem? –  Jonas Meyer Jan 16 '13 at 4:24
@JonasMeyer: We should be using l'hopital's rule since the limit reaches an indeterminate state at $0/0$, taking the derivative wrt $x$ should give us $\lim_{x \to 0} g'(0)$. Since we know that $g$ is differentiable at $0$, we can claim that the limit exists. Am I headed the right direction? –  user43901 Jan 16 '13 at 4:26
@user43901: You could do that if $g$ were assumed differentiable away from $0$, but it is not. Even if that were given, I'm biased against it in such a case. What is the definition of the derivative of $g$ at $0$? I don't know what you meant by "$\lim_{x\to 0}g'(0)$", but I'm guessing based on the statement that you wanted to use l'Hôpital's rule that you meant "$g'(x)$", which is a problem because it isn't assumed to exist. –  Jonas Meyer Jan 16 '13 at 4:27
@JonasMeyer: the derivative, according to the limit definition, is $g'(0)= \lim_{h \to 0} \frac {g(h)}{h}$. Is that what you were getting at? –  user43901 Jan 16 '13 at 4:30

Let $g(x)$ be a function which is differentiable at $0$ with $g(0)=0$. I claim that $f(x)=g(x)/x$ is continuous at $0$. It is easy to see that $f(x)$ is continuous around $0$ as the quotient of continuous functions, and we can define $f(0)=\lim_{x\to0}\frac{g(x)}{x}=g'(0)$ to make it continuous at $x=0$ by the definition of the derivative. Now, clearly $$g(x)=x\cdot\frac{g(x)}{x}=xf(x).$$

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I am failing to understand what follows your "now clearly"-- would you mind hashing this out a bit for me? –  user43901 Jan 16 '13 at 4:23
By using l'Hospital's rule, you agree $f(x)$ is continuous at $0$? If so, then all I am saying is that $x\cdot g(x)/x=(x/x)\cdot g(x)=1\cdot g(x)=g(x)$. –  Clayton Jan 16 '13 at 4:25
It is not assumed in the statement of the problem that $g'(x)$ exists when $x\neq 0$. –  Jonas Meyer Jan 16 '13 at 4:26
Since $g$ is differentiable at $0$, it is not only defined around an interval around $0$, it is even continuous (and differentiable) around $0$. –  Clayton Jan 16 '13 at 4:29
@Clayton: It is true that $g$ must be defined in an interval around $0$, but it need not even be continuous, let alone differentiable, away from $0$. For a discontinuous example, let $g(x)=x^2$ when $x$ is rational, $g(x)=0$ when $x$ is irrational. For a continuous example, let $f(x)$ be a continuous nowhere differentiable function, and let $g(x)=xf(x)$. On the other hand, defining $f(0)=\lim\limits_{x\to 0}\dfrac{g(x)}{x}$ works by the definition of the derivative and the assumption that $g$ is differentiable at $0$. –  Jonas Meyer Jan 16 '13 at 4:33