# Why is this piece-wise function differentiable at the point $x=0$?

I am given the following function:

$$f : \mathbb{R} \rightarrow \mathbb{R} \hspace{2cm}$$

$$f(x) = \left\{ \begin{array}{ll} x^2 \sin(\frac{1}{x}) & \quad x \neq 0 \\ 0 & \quad x = 0 \end{array} \right.$$

And I do not understand why this function is differentiable in $$x = 0$$, as my textbook claims. Firstly, I know that the function must be continuous in $$x=0$$ for us to even discuss the possibility of it being differentiable. For the function to be continuous at $$x=0$$, the limits of the function from both sides of $$x=0$$ must equal the value of the function itself at $$x=0$$. We have:

$$f(0)=0$$

$$\lim\limits_{x \to +0}f(x) = \lim\limits_{x \to +0} x^2 \sin\bigg (\frac{1}{x} \bigg ) = 0$$

$$\lim\limits_{x \to -0}f(x) = \lim\limits_{x \to -0} x^2 \sin\bigg (\frac{1}{x} \bigg ) = 0$$

So we can see that the function is indeed continuous at $$x=0$$. Now it is natural to discuss differentiability. For the function to be differentiable at $$x=0$$, the limit of the derivatives from both sides of $$x=0$$ must be equal to the derivative itself at $$x=0$$. If I find the derivative of $$f(x)$$ I get:

$$f'(x) = \left\{ \begin{array}{ll} 2x \sin(\frac{1}{x}) - \cos(\frac{1}{x}) & \quad x \neq 0 \\ 0 & \quad x = 0 \end{array} \right.$$

So

$$f'(0) = 0$$

$$\lim\limits_{x \to +0}f'(x) = \lim\limits_{x \to +0} 2x \sin \bigg( \frac{1}{x} \bigg) - \cos\bigg ( \frac{1}{x} \bigg )$$

$$\lim\limits_{x \to -0}f'(x) = \lim\limits_{x \to -0} 2x \sin \bigg( \frac{1}{x} \bigg) - \cos\bigg ( \frac{1}{x} \bigg )$$

And the last $$2$$ limits do not exist because of the $$\cos(\frac{1}{x})$$ term.

So I am really confused as to why this function is differentiable at $$x=0$$. What mistakes did I make in my reasoning? Why is the derivative of the function at $$x=0$$ well defined?

• While it is true that differentiability at $0$ implies continuity at $0$, that does not oblige you to first verify continuity at $0$ before you go on to consider differentiability at $0$. And, in fact, once you have demonstrated differentiability at $0$, you may conclude continuity at $0$ without any further work. That's how implications work. – Lee Mosher Dec 15 '19 at 1:32

"Differentiable at $$0$$" does not mean that the derivative is continuous, but rather that the derivative exists at $$0$$. You can see it by taking the limits of the Newton quotients at $$0$$: $$\frac{h^2\sin\frac1h}{h}=h\sin\tfrac1h\xrightarrow[h\to0]{}0,$$ so the derivative exists at $$0$$ and is $$0$$.

The "squeezing" by $$x^2$$ is what makes the function differentiable at $$0$$:

Diffentiability of $$f$$ at $$0$$ doesn't guarantee the continuity of $$f'$$ at $$0$$, and this is a classic example.

Indeed, the reasoning of the differentiability of $$f$$ at $$0$$ is the following.

For $$h\ne 0$$, $$|f(h)-f(0)|=|f(h)|=|h^{2}\sin(1/h)|$$, so $$|f(h)/h|\leq|h||\sin(1/h)|\leq|h|\rightarrow 0$$, therefore $$f'(0)=0$$.