도함수의 뜻

The function is differentiable at $x=1$ if the following limit: $$\lim_{x\to 1}\frac{f(x)-f(1)}{x-1} = \lim_{x\to 1}\frac{f(x)-0}{x-1}=\lim_{x\to 1} \frac{f(x)}{x-1}$$ exists (note that we used the fact that $f(1)=0$ as per the definition of the first branch which includes the point $x=0$). We compute left and right limits; so $$\lim_{x\to 1^-}\frac{f(x)}{x-1}=\lim_{x\to 1^-}\frac{4x^2-8x+4}{x-1}=\lim_{x\rightarrow 1^-}\frac{4(x-1)^2}{x-1} =\lim_{x\to 1^-} 4(x-1)=0$$ and $$\lim_{x\to 1^+}\frac{(x-1)^2\sin\left(\frac{1}{x-1}\right)}{x-1}=\lim_{x\to 1^+}(x-1)\cdot \sin\left(\frac{1}{x-1}\right).$$ This last limit equals $0$ by the Squeeze Theorem since $$-1\le \sin\left(\frac{1}{x-1}\right)\le 1$$ and so, $$-(x-1)\le (x-1)\cdot \sin\left(\frac{1}{x-1}\right)\le x-1,$$ where in these inequalities we used the fact that $x\to 1^+$ yields positive values for $x-1$. Finally, since $\lim_{x\to 1^+}-x+1=\lim_{x\to 1^+}x-1=0$, the Squeeze Theorem yields that also $\lim_{x\to 1^+}(x-1)\sin\left(\frac{1}{x-1}\right) =0$, as claimed.

Since the left and right limits match (they're both equal to $0$), we conclude that indeed $f(x)$ is differentiable at $x=1$ (and its derivative at $x=1$ is actually equal to $0$).