Powers Drown Logarithms
From ProofWiki
Contents |
Theorem
Let $r \in \Q_{>0}$ be a strictly positive rational number.
Can't this be extended to $\R_{>0}$ for free?
You can help Proof Wiki by completing it.
To discuss it in more detail, feel free to use the talk page.
When this has been done, {{WIP}} should be removed from the code.
If you would welcome a second opinion as to whether your work is correct, add a call to {{Proofread}} the page (see the proofread template for usage).
Then:
- $(1): \quad \displaystyle \lim_{x \to \infty} x^{-r} \ln x = 0$
- $(2): \quad \displaystyle \lim_{y \to 0_+} y^r \ln y = 0$
Proof
Proof of First Result
From Upper Bound of Natural Logarithm:
When $x > 1$:
- $\forall s \in \R: s > 0: \ln x \le \dfrac {x^s} s$
Given that $r > 0$, we can plug $s = \dfrac r 2$ in:
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle x^{-r} \ln x\) | \(=\) | \(\displaystyle x^{-r/2} \left({x^{-s} \ln x}\right)\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\le\) | \(\displaystyle \frac {x^{-r/2} } s\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | |||
| \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) | \(=\) | \(\displaystyle s \frac 1 {x^{r/2} }\) | \(\displaystyle \) | \(\displaystyle \) | \(\displaystyle \) |
From Power of Reciprocal:
- $\displaystyle \lim_{x \to \infty} x^{-r} \frac 1 {x^{r/2}} = 0$
and so:
- $\displaystyle \lim_{x \to \infty} x^{-r} \ln x = 0$
by the Squeeze Theorem.
$\blacksquare$
Proof of Second Result
Put $y = \dfrac 1 x$ in the first result.
$\blacksquare$
Sources
- K.G. Binmore: Mathematical Analysis: A Straightforward Approach (1977)... (previous)... (next): $S 14.3 \ (2) \ \text{(i), (ii)}$
