Real Natural Logarithm Function is Continuous
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Theorem
The real natural logarithm function is continuous.
Proof 1
We have that the Natural Logarithm Function is Differentiable.
The result follows from Differentiable Function is Continuous.
$\blacksquare$
Proof 2
From Bounds of Natural Logarithm:
- $\dfrac 1 2 < \map \ln 2 < 1$
Fix $x \in \R$.
Consider $\dfrac x {\map \ln 2}$.
From Rationals are Everywhere Dense in Topological Space of Reals:
- $\forall \epsilon \in \R_{>0} \exists r \in \Q : \size {r - \dfrac x {\map \ln 2} } < \epsilon$
Thus:
\(\ds \size {r - \dfrac x {\map \ln 2} }\) | \(<\) | \(\ds \epsilon\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \map \ln 2 \size {r - \dfrac x {\map \ln 2} }\) | \(=\) | \(\ds \size {\map \ln {2^r} - x }\) | Natural Logarithm of Rational Power | ||||||||||
\(\ds \) | \(<\) | \(\ds \epsilon \, \map \ln 2\) | Real Number Ordering is Compatible with Multiplication | |||||||||||
\(\ds \) | \(<\) | \(\ds \epsilon\) | as $\map \ln 2 < 1$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \size {\map \ln t - x}\) | \(<\) | \(\ds \epsilon\) | substituting $t = 2^r$ |
Thus:
- $\forall \epsilon \in \R_{>0}: \exists t \in \R_{>0}: \size {\map \ln t - x} < \epsilon$
Thus, the image of $\R_{>0}$ under $\ln$ is everywhere dense in $\R$.
From Monotone Real Function with Everywhere Dense Image is Continuous, $\ln$ is continuous on $\R_{>0}$.
Hence the result.
$\blacksquare$