$-\ln{x}<\frac{1}{x}$ when $0<x<1$

Question

I am trying to prove this inequality as described in the title:

$0<-\ln{x}<\frac{1}{x}$ when $0<x<1$

Graphing it will tell you this inequality is true. Here are my attempts and I could not produce a clean proof without graphing.

\begin{align*} \lim_{x \to 0^+} x \ln x = \lim_{x \to 0^+} \frac{\ln x}{1/x} = \lim_{x \to 0^+} \frac{1/x}{-1/x^2} = 0 \end{align*}

this shows that the inequality is true for at least some small domain $D=(0,\delta)$ but I could not stretch this $\delta$ precisely.

\begin{align*} -\ln{x}<\frac{1}{x} &\iff \ln{x}>-\frac{1}{x} \\ &\iff x > e^{-\frac{1}{x}} \end{align*} and I am stuck. The LHS tends to 0 and RHS tends to e^-infinity, which is also 0.

Any help would be appreciated, thanks.

Answer 1

$\ln y <y$ for $y>1$. To see this note that $(\ln y-y)'=\frac 1y -1 <0$ so $\ln y-y$ is a decreasing function which has the value $-1$ when $y =1$.

Now take $y=\frac 1 x$ to see that $-\ln x <\frac 1 x $.