\(\ln x\leqslant x-1\)(证明略)
\((1)\)当\(a\leqslant 0\)时，$a\ln x\geqslant a(x-1)\Rightarrow x-1\geqslant a(x-1)\Rightarrow\left{
\begin{array}{ll}
x>1 \
a\leqslant 1
\end{array}
\right.\(且\)\left{
\begin{array}{ll}
0<x<1\
a\geqslant 1
\end{array}
\right.\Rightarrow a=1\(与\)a\leqslant 0$矛盾，此时无解；
\((2)\)当\(a> 0\)时，\(a\ln x\leqslant a(x-1)\)
$1^\circ x-1\leqslant a(x-1)\Rightarrow\left{
\begin{array}{ll}
x>1 \
a\geqslant 1
\end{array}
\right.\(且\)\left{
\begin{array}{ll}
0<x<1\
a\leqslant 1
\end{array}
\right.\Rightarrow a=1$；
$2^\circ x-1\geqslant a(x-1)\Rightarrow\left{
\begin{array}{ll}
x>1 \
a\leqslant 1
\end{array}
\right.\(且\)\left{
\begin{array}{ll}
0<x<1\
a\geqslant 1
\end{array}
\right.\Rightarrow a=1$；
\(a=1.\)
\(a=1\)
\(\ln x\leqslant x-1\)
\(g(x)\geqslant 0\Rightarrow \ln x\leqslant x-1.\)
\(.\)
\(f(x)\geqslant 0\Rightarrow x-1\geqslant a\ln x\)
\(g(x)=\dfrac{x-1}{\ln x}\Rightarrow g'(x)=\dfrac{\ln x-1-\frac{1}{x}}{\ln^2 x}=\dfrac{h(x)}{\ln^2 x}\)
\(\Rightarrow h'(x)=x-\dfrac{1}{x^2}=\dfrac{x-1}{x^2}\Rightarrow h(x)\)在\((0\)，\(1)\)上单调递减，在\((1\)，\(+\infty)\)上单调递增，
\(\Rightarrow h(x)\geqslant 0\Rightarrow g'(x)\geqslant 0\Rightarrow g(x)\)在\((0\)，\(1)\)上单调递增，在\((1\)，\(+\infty)\)上单调递增，
\(\lim\limits_{x \to 1 }\dfrac{x-1}{\ln x}=\lim\limits_{x \to 1 }\dfrac{1}{\frac{1}{x}}=1\)
\(a\leqslant \dfrac{x-1}{\ln x}\to 1(x\to 1)\)；
\(a\geqslant \dfrac{x-1}{\ln x}\to 1(x\to 1)\)；
\(a=1.\)