$A propriety of a given fuction

Question

Let $f \colon [0; 1] \to \mathbb{R}$ be given by $$ f(x) = \begin{cases} e^{x} & x \in (0, 1]\\ 0 & x= 1 \end{cases} $$

Is it true that $f$ is continuous on $[0, 1]$?

Answer 1

HINT: Check that $f$ is continuous. Then it is easy to observe, that its derivative is bounded. Indeed: $$ f'(x)=-e^{-1/x}\frac1{x^2}=-\frac{1}{x^2e^{1/x}}=-\frac{(1/x)^2}{e^{1/x}}, $$ which tends to 0, when $x\to0+$. Hence the derivative is bounded on $(0,1]$, so $f$ is absolutely continuous on $(0,1]$ and, as a result, on $[0,1]$.

Remark: If for somebody it is not a simple conclusion of continuity at $0$: is it obvious that $e^{-x}$ is absolutely continuous on $(-\infty,0)$?