Lower Bound of Natural Logarithm/Proof 1
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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