Justify if $fg$ is differentiable at point $x_0=0$

Question

I had these two statements in some exam recently and I still wonder how to tackle them to decide which one is correct.

1. $f$ is continuous at $0$, $g$ is differentiable at $0$, $g(0)=0$, $fg$ is differentiable at $0$.

2. $f$ is continuous at $0$, $g$ is differentiable at $0$, $f(0)=0$, $fg$ is differentiable at $0$.

I know that if $g$ is differentiable at a certain point $x_0$, then $g$ must also be continuous at $x_0$. Also, $f$ is differentiable at $x_0$ iff $\lim \limits_{x \to x_0 } \frac{f(x) - f(x_0)}{x - x_0}$ exists and that value is denoted by $f'(x_0)$. But I'm confused about the importance of $g(0)=0$ and $f(0)=0$.

Answer 1

I'm confused about the importance of $g(0)=0$ and $f(0)=0$.

Here's the idea of why that's important. Suppose we try applying the product rule. Then

$$ (fg)'(0) = f(0)g'(0) + f'(0)g(0) $$

Now we don't know that $f'(0)$ exists so if $g(0) = 0$ then we "don't have to worry about the last term." That is, if $g(0) = 0$ we might expect

$$ (fg)'(0) = f(0)g'(0). $$

Answer 2

Hint:

1. define the function $h$ as $h = fg$.
2. Write down, by definition, what it means for $h$ to be differentiable at $0$.
3. Write down, by definition, what it means for $g$ to be differentiable at $0$.
4. Compare the two expressions from $2$ and $3$.

What did you get as the result of step $2$? What did you get as the result of step $3$? How do the two compare?

Answer 3

I will give you a "proof" that is a bit harder than you might hope, but I invite you to think about it.

Since $g$ is differentiable at zero, we can write $$ g(x)=g(0)+g'(0)x + o(x) \tag{1} $$ as $x \to 0$. Here $\lim_{x \to 0} \frac{o(x)}{x}=0$. Now, $$ f(x)g(x) = f(x) \left( g(0)+g'(0)x+o(x) \right) = f(x)g(0) + g'(0)f(x)x + f(x)o(x). \tag{2} $$ We must decide whether the right-hand side of (2) is an affine function plus $o(x)$ as $x \to 0$. But $$ \lim_{x \to 0} \frac{f(x)g(x)}{x} = \lim_{x \to 0} \frac{f(x)g(0) + g'(0)f(x)x + f(x)o(x)}{x}. $$ It follows easily that if $g(0)=0$, then $$ \lim_{x \to 0} \frac{f(x)g(x)}{x} = g'(0)f(0) $$ due to the continuity of $f$. The correct answer should be 1.