Logarithm Tends to Infinity

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

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

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

Proof 1
From Natural Logarithm of 2 Greater than One Half, we have that:
 * $\ln 2 \ge \dfrac 1 2$

From the definition of infinite limits at infinity, our assertion is:


 * $\forall M > 0 : \exists N > 0 : x > N \implies \ln x > M$.

As $x \to +\infty$, we will restrict our attention to sufficiently large $M$.

Now, because the natural logarithm is increasing, for sufficiently large $M$:


 * $x > 2^{2M} \implies \ln x > \ln 2^{2N}$

From the Laws of Logarithms:


 * $\quad \ge 2M \cdot \dfrac 1 2 = M$

Choosing $N = \ln 2^{2M}$:


 * $\forall M \ge a: \exists N > 0: x > N \implies \ln x > M $

where $a$ is some large number.

Hence the result, by the definition of infinite limits at infinity.

Proof 2
From the definition of natural logarithm (or from Equivalence of Logarithm Definitions):

The result follows from Integral of Reciprocal is Divergent.

Also see

 * Logarithm Tends to Negative Infinity