# Checking differentiability of a piece-wise defined function using the limit of the left and right derivative

$$f=\begin{cases}\cos(x+\pi) &,x>0\\x^2+x-1 &,x\leq 0\end{cases} \tag{1}$$

Problem 1: Where's $$f$$ continuous?

$$f$$ is continuous on $$\mathbb R\setminus\{0\}$$ since it consists of continuous functions there. We check $$x=0$$.

$$\lim_{x\to 0^{+}}f(x)=\lim_{x\to 0^{+}}\cos(x+\pi)=\cos(\pi)=-1 \tag{2}$$ and $$\lim_{x\to 0^{-}}f(x)=\lim_{x\to 0^{-}} x^2+x-1=-1 \tag{3}$$

so $$f$$ is continuous on $$\mathbb R$$.

Problem 2: Where is $$f$$ differentiable?

The only problem point might be $$x=0$$. We have $$f'(x)=\begin{cases}-\sin(x+\pi) &, x>0\\ 2x+1 &, x<0\end{cases}\tag{4}$$ so f is diferentiable on $$\mathbb R\setminus\{0\}$$.

I'd like to use two different approaches here:

Approach 1: Definition

$$\lim_{h\to 0^{+}}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0^{+}}\frac{\cos(h+\pi)-(-1)}{h}=\lim_{h\to 0^{+}}\frac{-\sin(h+\pi)}{1}=0 \tag{5}$$

$$\lim_{h\to 0^{-}}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0^{-}}\frac{h^2+h-1-(-1)}{h}=\lim_{h\to 0^{+}}\frac{h^2+1}{1}=1 \tag{6}$$

so we can conclude, that $$f$$ is not differentiable at $$x=0$$

Approach 2: Limit

We see

$$\lim_{x\to 0, x>0} f'(0)=0, \quad \lim_{x\to 0, x<0} f'(0)=1 \tag{7}$$

so $$f$$ is not differentiable at $$x=0$$

Question: I think I misunderstand approach 2 sind I am not sure how approach 2 shows that $$f$$ is not differentiable at $$x=0$$. It'd make sense that $$f$$ is differentiable if those limits would've been the same but since they aren't, isn't the only statement we can make that $$f$$ isn't continuous differentiable at $$x=0$$? Couldn't it be, that $$f'$$ might be discontinuous at $$x=0$$ but still differentiable?

Consider the function $$f:\mathbb{R} \to \mathbb{R}$$ defined as $$f(x)= \begin{cases} x^2\sin\left(\dfrac1x\right), \ x \neq 0\\ 0, \ x=0. \end{cases}$$
This function is differentiable everywhere. At $$x=0$$ we just calculate $$\lim_{h \to 0} \frac{f(h)-f(0)}{h} = \lim_{h \to 0} \frac{h^2\sin(1/h)}{h} = \lim_{h \to 0} h\sin(1/h) = 0.$$
At any other point, we can obtain the derivative by the usual rules of differentiation. Therefore, we have $$f'(x)= \begin{cases} 2x\sin(1/x)-\cos(1/x), \ x\neq 0\\ 0, \ x=0. \end{cases}$$
Notice that $$\lim_{x \to 0}f'(x)$$ does not exist, so even though $$f$$ is differentiable everywhere, the derivative is not continuous at $$x=0$$.