Prove a piecewise function is differentiable at $ x = 1 $

Question

Use the difference quotient definition of the derivative, $$ f ' ( a ) = \lim _ { x \to a } \frac { f ( x ) - f ( a ) } { x - a } \text , $$ to show that $$ f ( x ) = \begin {cases} x \ln x \text , & x \le 1 \\ x - 1 \text , & x > 1 \end {cases} $$ is differentiable at $ x = 1 $.

Answer 1

Note that on the one hand $$ f'_{+}(a) =\lim_{x\to1^{+}}\frac{\overbrace{f(x)}^{x-1}-\overbrace{f(1)}^{0}}{x-1} =\lim_{x\to1^{+}}\frac{x-1}{x-1}=1 $$ On the other hand $$ \begin{split} f'_{-}(a) &=\lim_{x\to1^{-}}\frac{\overbrace{f(x)}^{x\ln x}-\overbrace{f(1)}^{0}}{x-1} =\lim_{x\to1^{-}}\frac{x\ln x}{x-1}\\ &\overset{\small t=x-1}{=} \lim_{t\to0^{-}}\frac{(t+1)\ln(t+1)}{t}\\ &=\lim_{t\to0^{-}}(t+1)\lim_{t\to0^{-}}\frac{\ln(t+1)}{t}\\ &=1\cdot(0+1)=1 \end{split} $$ Therefore, $f'_{+}(1)=f'_{-}(1)$, so $f'(1)=1$, as required.