Differentiability of $g(x)\sin(1/x)$ at $x=0$.

Question

I have a solution to the following exercise, and would like to confirm if I am proceeding correctly:

Suppose that $g(x)$ is differentiable at $x=0$, and that $g(0)=g'(0)=0$. Show that $$f(x)=\begin{cases} g(x)\sin(x) && x\neq 0 \\ 0 && x=0\end{cases}$$ is differentiable at $x=0$.

My attempt: To show that $f(x)$ is differentiable at $x=0$, we would like to show that $\lim_{h\to 0}\frac{f(h)-f(0)}{h}$ exists. We have that \begin{align*} \lim_{h\to 0}\frac{f(h)-f(0)}{h} & = \lim_{h\to 0} \frac{g(h)\sin(1/h)-0}{h} \\ & = \lim_{h\to 0}\frac{g(h)}{h}\cdot\sin(1/h) \end{align*} Since $g'(0)=0$, we have that $\lim_{h\to 0}\frac{g(h)-g(0)}{h}=\lim_{h\to 0}\frac{g(h)}{h}=0$. Therefore, $$\lim_{h\to 0}\frac{g(h)}{h}\cdot\sin(1/h)=0\cdot\lim_{h\to 0}\sin(1/h)=0$$ Hence, $f'(0)=0$.

I am unsure whether I can say that $0\cdot\lim_{h\to 0}\sin(1/h)=0$.

Answer 1

The limit $\lim_{h\to 0}\sin(1/h)$ does not exist !

But since $|\frac{g(h)}{h}\cdot\sin(1/h)| \le |\frac{g(h)}{h}|$ and $\lim_{h\to 0}\frac{g(h)}{h}=0$, we see that $\lim_{h\to 0}\frac{f(h)-f(0)}{h}=0$.

Answer 2

Yes, you can say that $0\times\lim_{h\to0}\sin\left(\frac1h\right)=0$, but that's not a sensible thing to say.

It is better to say that$$\left\lvert\frac{g(h)}h\sin\left(\frac1h\right)\right\rvert\leqslant\left\lvert\frac{g(h)}h\right\rvert$$and then to uset the fact that$$\lim_{h\to0}\frac{g(h)}h=0\iff\lim_{h\to0}\left\lvert\frac{g(h)}h\right\rvert=0.$$

Answer 3

By Taylor's theorem: $$g(x)=g(0)+g'(0)x+r(x)x$$ $$g(x)=xr(x)$$ So $$f(x)=xr(x)\sin\left(\frac{1}{x}\right)$$ And $$f'(0)=\lim_{x \to 0} \frac{f(x)-f(0)}{x}=\lim_{x \to 0} r(x) \sin\left(\frac{1}{x}\right)=0$$ Because $r(x) \to 0$ and $\sin\left(\frac{1}{x}\right)$ is bounded.

Answer 4

If I am correct then the question mention's about

$$ f(x) = \begin{cases}0 & x = 0\\ g(x)\sin{x} & x \neq0\end{cases} $$

So I would use the following:

$$ f^{'}(x) = \lim_{h \rightarrow 0} \frac{f(x+h)-f(x)}{h} $$

$$f^{'}(0) = \lim_{h \rightarrow 0}\frac{f(h)-f(0)}{h} \\ = \lim_{h \rightarrow 0} \frac{g(h)\sin{h}-0}{h}$$

Now using the identity $ \lim_{h \rightarrow 0} \frac{\sin{h}}{h} = 1$.

This shortened the expression to

$$ \lim_{h \rightarrow 0} g(h) $$

Now note that $g$ is continuous about $x=0$, the function $f$ is differentiable about $x=0$.