# 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.

$\blacksquare$