Natural Logarithm Function is Continuous

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Theorem

The 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 Reals:

$\forall \epsilon \in \R_{>0} \exists r \in \Q : \size {r - \dfrac x {\map \ln 2} } < \epsilon$

Thus:

\(\displaystyle \size {r - \dfrac x {\map \ln 2} }\) \(<\) \(\displaystyle \epsilon\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \map \ln 2 \size {r - \dfrac x {\map \ln 2} }\) \(=\) \(\displaystyle \size {\map \ln {2^r} - x }\) Natural Logarithm of Rational Power
\(\displaystyle \) \(<\) \(\displaystyle \epsilon \, \map \ln 2\) Real Number Ordering is Compatible with Multiplication
\(\displaystyle \) \(<\) \(\displaystyle \epsilon\) as $\map \ln 2 < 1$
\(\displaystyle \leadsto \ \ \) \(\displaystyle \size {\map \ln t - x}\) \(<\) \(\displaystyle \epsilon\) 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.

$\blacksquare$