Proving a function is differentiable using the definition of a derivative. Let $ g:\mathbb{R}\to\mathbb{R} $ be a bounded function and define $ f:\mathbb{R}\to\mathbb{R} $ by $ f(x)=x^2g(x) $. Prove that $ f $ is differentiable at $ x = 0 $ and find the derivative of $ f $ at $ x=0 $.
Hello. I am a first year student thrown into real analysis and I'm not the best at writing proof. Down below is what I have arrived at. I don't know if this is enough though.
$\lim_{x\to x_0} \frac{f(x) - f(x_0)}{x-x_0}$ = the derivative
$f(x) = x^2g(x)$
$x_0 = 0$
$\lim_{x\to 0} \frac{x^2g(x) - f(0)}{x-0}$
$\lim_{x\to 0} \frac{0^2g(0) - 0^2g(0)}{0-0} = 0/0$
$f$ is differentiable at $x=0$ and the derivative is $0$.
Thank you.
 A: You were doing fine until that last line... but that last line is a bit of a doozy. :-)
You can't just plug in $x=0$ in a limit as $x\to0$ unless you know that the function you're taking the limit of is continuous.  Which is a big if.
So instead, focus on the limit that you've written down:
$$
\lim_{x\to0}\frac{f(x)-f(0)}{x-0}=\lim_{x\to0}\frac{x^2g(x)-0}{x-0}=\lim_{x\to0}xg(x).
$$
To show that the function is differentiable, you need to show that this limit exists.
To do that, I suggest you first think about what the limit "should" be. In this case, we're multiplying $x$ (which we know goes to $0$) by $g(x)$. What would $g(x)$ need to look like in order to make $xg(x)\nrightarrow0$?  How can the assumptions you were given about $g(x)$ help ensure that doesn't happen?
A: There are a lot of ways to get this question wrong.
You can't just plug in $x = 0$ into the limit, because you get the indeterminate form $0/0$. You can't use L'hopital's rule, but that requires you to already know the derivative, and $g$ might not be differentiable.
You might think we can do (this is a false proof)
$$\lim_{x \rightarrow 0} \frac{x^2 g(x)}{x} = \lim_{x \rightarrow 0} x g(x) =  \lim_{x \rightarrow 0}g(x) \cdot  \lim_{x \rightarrow 0} x = 0$$
But that's not correct, $g$ might not be continuous and $\lim_{x \rightarrow 0} g(x)$ might not even exist. Note that we haven't used the condition that $g$ is bounded in the false proof. The limit of a product is the product of the limits, but  the limits need to exist.
Instead, use the epsilon-delta definition directly. We want to compute $\lim_{x \rightarrow 0} x g(x)$. Let $\epsilon > 0$. Then, since $g$ is bounded, there is a $M \in \mathbb{R}$ such that for all $x$, $|g(x)| < M$.
Can you finish the proof from here?
