How can I prove that $ \frac {x-1}{x}\leq \log x\leq x-1$ How can I prove that $$ \frac {x-1}{x}\leq \log x\leq x-1$$
for any $x>1$
 A: The inequalities are tight for $x=1$.
Then for $x>1$, deriving,
$$\frac1{x^2}\le\frac1x\le1$$ holds.
A: I will do one inequality and you can do the other in a similar way.
Let $f(x)=\log x -\frac{x-1}{x}=\log x-1+\frac{1}{x}$. Note that $f(1)=0$ and $f'(x)=\frac{1}{x}-\frac{1}{x^2}$. So $f'(x)\gt 0$ if $x\gt 1$, the function $f$ is increasing.
A: A sketch: $\log x$ is concave, so its graph lies below its tangent line at any point, including $(1,0)$. This gives the right-hand inequality. The left-hand inequality follows from the same fact, after a variable change $x \rightarrow {1 \over x}$.
A: For $\frac{x-1}{x}\leq \log x$, denote $f(x)=\log x-\frac{x-1}{x}$. You have $f(1)=0$. 
$$f'(x)=\frac{1}{x}-\frac{1}{x^2}\leq 0\Leftrightarrow x\in [1,\infty)$$ and so $f$ is increasing. For $$\log x\leq x-1$$ again if we denote $g(x)=x-1-\log x$ we have $g(1)=0$ and $g'(x)\ge 0$ for $x\ge 1$
A: For any $x > 1$, the mean value theorem gives
$$\inf_{c \in (1,x)} \frac{1}{c} (x-1) \leq \log(x) - \log(1) \leq \sup_{c \in (1,x)} \frac{1}{c} (x-1)$$
$$\frac{x-1}{x} \leq \log(x) \leq x -1 . $$
A: $\log x = \int_{1}^{x} \frac{1}{x} dx$
$\frac{1}{x} < 1$ for all $x>1$
therefore, $\log x = \int_{1}^{x} \frac{1}{x} dx <  \int_{1}^{x} 1 dx = x-1$
for the left side:
note that $\frac{1}{x^2} < \frac{1}{x}$ for all $x>1$
