How to show that $\frac{\ln x}{x}$ is monotone for $x\ge e$?

Question

How to show that $\frac{\ln x}{x}$ is monotone for $x\ge e$?

Looking at the graph of $\ln x$ I can tell that for $x<e$ the $\ln x$ goes to $-\infty$ very fast and for $x\ge e$ it grows very slow.

Also, I know that $\ln x$ is monotone on $[0,\infty)$ and so is $g(x)=x$.

Answer 1

To check monotonicity you need to compute the sign of the first derivative.

Let $f(x)=\frac{\ln(x)}{x}$. Then

$$f'(x)=\frac{1-\ln(x)}{x^2}.$$

If $x\ge e$, then $\frac{1-\ln(x)}{x^2}\le \frac{1-\ln(e)}{x^2}=0$, so $f$ is monotonously decreasing for $x\ge e$.

Answer 2

HINT: you have to consider $$y'=\frac{1-\ln(x)}{x^2}$$

Answer 3

A function is monotone on an interval $[a,b]$ if its derivative is either $\leq0$ (monotonously decreasing) or $\geq0$ (monotonously increasing) on the whole interval. Thus, differentiate ${\ln x \over x}$:

$$\left({\ln x \over x}\right)' = {1 - \ln x \over x^2}$$

Now, show that ${\ln x \over x}$ is monotonously decreasing on $[e, \infty)$: $${1 - \ln x \over x^2} \leq 0 \text{ } \forall x \geq e$$

$$\iff 1 \leq x^2 + \ln x $$

Which is obviously true for $x \geq e$.