Limit of x to the x/Proof 2

Theorem

Let $f: \R_{>0} \to \R$ be defined as:

$\forall x \in \R_{>0}: \map f x = x^x$

Then:

$\ds \lim_{x \mathop \to 0^+} x^x = 1$

Proof

\(\ds \lim_{x \mathop \to 0^+} x^x\)

\(=\)

\(\ds \lim_{x \mathop \to 0^+} \map \exp {x \ln x}\)

Definition 1 of Power (Algebra)

\(\ds \)

\(=\)

\(\ds \map \exp {\lim_{x \mathop \to 0^+} x \ln x}\)

Exponential Function is Continuous

\(\ds \)

\(=\)

\(\ds \exp 0\)

Limit of Power of $x$ by Absolute Value of Power of Logarithm of $x$: Corollary

\(\ds \)

\(=\)

\(\ds 1\)

Exponential of Zero

$\blacksquare$