Properties of Natural Logarithm

Theorem
Let $x \in \R$ be a real number such that $x > 0$.

Let $\ln x$ be the natural logarithm of $x$.

Then:
 * $\ln 1 = 0$;
 * $D \ln x = \frac 1 x$
 * The function $f \left({x}\right) = \ln x: x > 0$ is strictly increasing and concave;
 * $\ln x \to +\infty$ as $x \to +\infty$;
 * $\ln x \to -\infty$ as $x \to 0^+$.

Proof

 * $\ln 1 = 0$:

From the definition, $\displaystyle \ln x = \int_1^x \frac {dt} t$.

From Integral on Zero Interval, $\displaystyle \ln 1 = \int_1^1 \frac {dt} t = 0$.


 * $D \ln x = \frac 1 x$

From the definition, $\displaystyle \ln x = \int_1^x \frac {dt} t$.

Thus from the Fundamental Theorem of Calculus we have that $\ln x$ is a primitive of $\displaystyle \frac 1 x$ and hence $\displaystyle D \ln x = \frac 1 x$.


 * $\ln x$ is strictly increasing and concave:

From the above $\displaystyle D \ln x = \frac 1 x$, which is strictly positive on $x > 0$.

From Derivative of Monotone Function it follows that $\ln x$ is strictly increasing on $x > 0$.

From the Power Rule for Derivatives: Integer Index, $\displaystyle D^2 \ln x = D \frac 1 x = \frac {-1} {x^2}$.

Thus $D^2 \ln x$ is strictly negative on $x > 0$ (in fact is strictly negative for all $x \ne 0$).

Thus from Derivative of Monotone Function, $\displaystyle D \frac 1 x$ is strictly decreasing on $x > 0$.

So from Derivative of Convex or Concave Function, $\ln x$ is concave on $x > 0$.


 * $\ln x \to +\infty$ as $x \to +\infty$:

This follows directly from Integral of Reciprocal is Divergent.


 * $\ln x \to -\infty$ as $x \to 0^+$.

This follows directly from Integral of Reciprocal is Divergent.