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:
 * $(1): \quad \ln 1 = 0$
 * $(2): \quad D \ln x = \dfrac 1 x$
 * $(3): \quad \ln x: x > 0$ is strictly increasing and concave
 * $(4): \quad \ln x \to +\infty$ as $x \to +\infty$
 * $(5): \quad \ln x \to -\infty$ as $x \to 0^+$.

Proof
$(1): \quad \ln 1 = 0$:

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

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

$(2): \quad D \ln x = \dfrac 1 x$:

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 $\dfrac 1 x$ and hence $D \ln x = \dfrac 1 x$.

$(3): \quad \ln x$ is strictly increasing and concave:

From the above $D \ln x = \dfrac 1 x$, 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, $D^2 \ln x = D \dfrac 1 x = \dfrac {-1} {x^2}$.

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, $D \dfrac 1 x$ is strictly decreasing on $x > 0$.

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

$(4): \quad \ln x \to +\infty$ as $x \to +\infty$:

This follows directly from Integral of Reciprocal is Divergent.

$(5): \quad \ln x \to -\infty$ as $x \to 0^+$:

This follows directly from Integral of Reciprocal is Divergent.