Limit of x to the x
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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 1
| \(\ds \lim_{x \mathop \to 0^+} x^x\) | \(=\) | \(\ds \lim_{x \mathop \to 0^+} \map \exp {x \ln x}\) | Definition 1 of Power (Algebra) | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \exp {\lim_{x \mathop \to 0^+} x \ln x}\) | Exponential Function is Continuous | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \exp {\lim_{x \mathop \to 0^+} \frac {\ln x} {\frac 1 x} }\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \map \exp {\lim_{x \mathop \to 0^+} \frac {\frac 1 x} {\frac {-1} {x^2} } }\) | L'Hôpital's Rule | |||||||||||
| \(\ds \) | \(=\) | \(\ds \map \exp {\lim_{x \mathop \to 0^+} \frac {-x^2} x}\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds \exp 0\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds 1\) | Exponential of Zero |
$\blacksquare$
Proof 2
| \(\ds \lim_{x \mathop \to 0^+} x^x\) | \(=\) | \(\ds \lim_{x \mathop \to 0^+} \map \exp {x \ln x}\) | 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 |
$\blacksquare$
Also presented as
This can also be presented as:
- $\ds \lim_{x \mathop \to 0^+} \map \exp {x \ln x} = 1$
as is demonstrated during the course of the proof.