# Proof that f(x) = ln(x) - x has a maximum, without using derivatives and taylor series expansion

#### Erradium

I need to proove that for any x in the range (x>0), f(x) <= f(1) = -1.
I tried using the limit definition of e, and the fact that e^x > x for any x>0, to no avail.
Can someone help me?

#### Prove It

MHF Helper
You wouldn't know that \displaystyle \begin{align*} \mathrm{e}^x > x \end{align*} without knowledge of Taylor Series, so I doubt that result could be used.

I really dislike when people put unnecessary restrictions on themselves. Always remember Occam's Razor - the most concise solution is the best one.

Plato

#### Idea

maybe we can use this

$$\displaystyle \ln x -x +1=\int_1^x \left(\frac{1}{t}-1\right) \, dt$$

#### Archie

You wouldn't know that \displaystyle \begin{align*} \mathrm{e}^x > x \end{align*} without knowledge of Taylor Series
Why not? I can prove $\displaystyle \lim_{x \to \infty} \frac{\ln x}{x} = 0$ without recourse to Taylor series. On the other hand, I'd use the integral definition of the logarithm which rather suggests knowledge of derivatives.

#### Prove It

MHF Helper
Why not? I can prove $\displaystyle \lim_{x \to \infty} \frac{\ln x}{x} = 0$ without recourse to Taylor series. On the other hand, I'd use the integral definition of the logarithm which rather suggests knowledge of derivatives.
Interesting, how would you do it?

#### Archie

\begin{align}
0 \lt \frac1t & \lt t^{c-1} &(t \gt 1, \, c \gt 0) \\
0 \lt \int_1^x \frac1t \,\mathrm dt & \lt \int_1^x t^{c-1} \,\mathrm dt \\
0 \lt \log t &\lt \frac1c ( x^c -1) \lt \frac{x^c}c \\
0 \lt \frac{\log^b x}{x^a} \lt \frac {x^{-a}}{c^b} &(a \gt 0, \, b \gt 0) \\
\text{(choose $c= \tfrac{a}{2b}$ and} \; 0 \lt \frac{\log^b x}{x^a} \lt \frac {x^{-\frac{a}2}}{c^b}
\end{align}
and take limits as $x \to \infty$.

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