16.已知函数$f(x) = \frac{\ln x - a}{e^{x - 1}}$，$g(x) = \frac{x - a - 1}{x}$，若对任意的$x \in \lbrack 1, + \infty )$，$f(x) \leq g(x)$恒成立，则实数$a$的取值范__．
解：依题可知$\frac{\ln x - a}{e^{x - 1}} \leq \frac{x - a - 1}{x}$（$x \geq 1$）恒成立
$\therefore x(\ln x - a) \leq e^{x - 1}(x - a - 1)$
$\therefore e^{x - 1}(\ln x - a) \leq e^{x - 1}(x - 1) - a$
设$t(x) = \ln x - x + 1$，则$t'(x) = \frac{1}{x} - 1 = \frac{1 - x}{x}$
$\therefore 0 < x < 1$时$t(x)$单增，$x > 1$时$t(x)$单减
$\therefore t(x)$极大值$= t(1) = 0$
$\therefore \ln x \leq x - 1$
设$h(x) = e^{x}(x - a)$，则$h(\ln x) \leq h(x - 1)$恒成立
$\therefore x \geq 0$时，$h(x)$单增
$\because h'(x) = e^{x}(x - a + 1)$
$\therefore x - a + 1 \leq 0(x \geq 0)$恒成立
$\therefore a \geq x + 1(x \geq 0)$恒成立
$\therefore a \leq 1$