Logarithm Tends to Infinity

Theorem

Let $x \in \R_{>0}$ be a strictly positive real number.

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 is Greater than One Half:

$\ln 2 \ge \dfrac 1 2$

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

$\forall M \in \R_{>0} : \exists N > 0 : x > N \implies \ln x > M$.

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

From Logarithm is Strictly Increasing:

$\ln x$ is strictly increasing.

So, for sufficiently large $M$:

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

From the Laws of Logarithms:

\(\ds \ln 2^{2 M}\)

\(=\)

\(\ds 2 M \ln 2\)

\(\ds \)

\(\ge\)

\(\ds 2 M \cdot \dfrac 1 2\)

\(\ds \)

\(=\)

\(\ds M\)

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

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

for some $a \in \R$.

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

$\blacksquare$

Proof 2

From the definition of the natural logarithm:

\(\ds \ln x\)

\(=\)

\(\ds \int_1^x \dfrac 1 t \rd t\)

The result follows from Integral of Reciprocal is Divergent.

$\blacksquare$

Also see

• Logarithm Tends to Negative Infinity

Sources

• 2005: Roland E. Larson, Robert P. Hostetler and Bruce H. Edwards: Calculus (8th ed.): Appendix $A$: Properties of the Natural Logarithmic Function