Prove the inequalities 1 - 1/x < log x < x - 1

Define the following functions:
$f(x) = x - 1 - \log x, \qquad g(x) = \log x - 1 + \frac{1}{x},$

for $x > 0$ . Prove that for $x > 0$ and $x \neq 1$ the inequalities

$1 - \frac{1}{x} < \log x < x - 1$

hold. Proceed by examining the signs of the derivatives $f'$ and $g'$ . When $x = 1$ these are equalities.
Draw the graphs of the functions
$A(x) = x - 1, \qquad B(x) = 1 - \frac{1}{x}, \qquad x > 0.$

Interpret the inequalities in part (a) geometrically.

First, we compute the derivatives of the functions,
$\begin{align*} f(x) &= x - 1 - \log x & \implies && f'(x) &= 1 - \frac{1}{x} \\ g(x) &= \log x - 1 + \frac{1}{x} & \implies && g'(x) &= \frac{1}{x} - \frac{1}{x^2} = \frac{1}{x} \left( 1 - \frac{1}{x} \right). \end{align*}$

Then, considering the derivative of $f$ ,

$\begin{align*} f'(x) = 1 - \frac{1}{x} && \implies && f'(x) > 0 \quad &\text{when } x > 1 \\ \text{and} && \implies && f'(x) < 0 \quad &\text{when } 0 < x < 1. \end{align*}$

Therefore, the function $f$ has a minimum at $x = 1$ . Since we can directly evaluate $f(1) = 0$ , this means $f(x) > 0$ for $x > 0$ and $x \neq 1$ . Therefore, $x - 1 > \log x$ .

Next, looking at the derivative of $g$ ,

$\begin{align*} g'(x) = \frac{1}{x} \left( 1 - \frac{1}{x} \right) && \implies && g'(x) > 0 \quad & \text{when } x > 1 \\ \text{and} && \implies && g'(x) < 0 \quad &\text{when } 0 < x < 1. \end{align*}$

Therefore, the function $g$ has a minimum at $x = 1$ . Since $g(1) = 0$ this implies $g(x) > 0$ for $x > 0$ and $x \neq 1$ . Thus,

$1 - \frac{1}{x} < \log x.$

Putting these two pieces together we have established the requested inequalities:

$1 - \frac{1}{x} < \log x < x - 1 \qquad \text{for } x > 0 \text{ and } x \neq 1. \qquad \blacksquare$
The graph of the two functions is:

The inequalities in part (a) imply that the graph of $\log x$ must lie strictly between the graphs of $A(x)$ and $B(x)$ shown above (and so $\log 1 = 0$ ).