Logarithm Tends to Infinity

From ProofWiki
Jump to navigation Jump to search

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:

\(\ds \ln x\) \(=\) \(\ds \int_1^x \dfrac 1 t \ \mathrm d t\)

The result follows from Integral of Reciprocal is Divergent.

$\blacksquare$


Also see


Sources