Can any one tell me if I did it correctly? Use the Mean Value Theorem to prove that  $(x-1)/x< In x< x-1$ for $x>1$. [Hint: use the fact D in $x=1/x$ for $x>0$.]
Can I say: for $(x-1)/x< In x$, we have $f(x)=Inx -(x-1)/x$, $f(1)=0$. $f'(x)={1}/{x}-1/x^2=1/x(1-1/x)>0$ for $x>0$. Thus, $f(x)$ is an increasing function for $x>1$. For $In x< x-1$ , we have $f(x)=(x-1)- Inx $, $f(1)=0$. $f'(x)=1-1/x$  for $x>0$  . Thus, $f(x)$ is an increasing function for $x>1$. Therefore, $(x-1)/x< In x< x-1$ for $x>1$.? Or any one has a better way to do it with the hint.

We know that
$$\frac{\ln x}{x-1}=\frac{\ln x-\ln 1}{x-1}\overset{\text{MVT}}=\ln'\xi=\frac1\xi$$
for some $\xi\in(1,x)$. But
\begin{align*}\frac1\xi<1&\implies\ln x<x-1\\\frac1\xi>\frac1x&\implies\ln x>\frac{x-1}x,\end{align*}
so we are done.
