Lower Bound of Natural Logarithm/Proof 1

Theorem

$\forall x \in \R_{>0}: 1 - \dfrac 1 x \le \ln x$

where $\ln x$ denotes the natural logarithm of $x$.

Proof

Let $x > 0$.

\(\ds x - 1\)

\(\ge\)

\(\ds \ln x\)

Upper Bound of Natural Logarithm

\(\ds \leadsto \ \ \)

\(\ds \frac 1 x -1\)

\(\ge\)

\(\ds \ln \frac 1 x\)

putting $\frac 1 x$ into the above inequality

\(\ds \)

\(=\)

\(\ds -\ln x\)

Logarithm of Reciprocal

\(\ds \leadsto \ \ \)

\(\ds 1 - \frac 1 x\)

\(\le\)

\(\ds \ln x\)

multiplying throughout by $-1$

$\blacksquare$

Illustration

Also see

• Bounds of Natural Logarithm