Is it true that $x^n <\epsilon \Rightarrow n < \frac{\ln \epsilon}{\ln x}$?

Question

Let:

1. $0 \lt x \lt 1$
2. $\epsilon > 0$

I need to show that there exists an $N(\epsilon,x)$ such that:

3. $n\ge N(\epsilon,x) \Rightarrow x^n < \epsilon$

This is what I've tried:

4. $x^n <\epsilon \Rightarrow \ln(x^n) <\ln(\epsilon) \Rightarrow n < \frac{\ln(\epsilon)}{\ln(x)}$

Now, when I let $\epsilon=1,x=0.5$, I get $n<0$. But if let $n=-1$, I get $x^n=0.5^{-1}=2$ which is more than $\epsilon=1$.

What did I do wrong? I also expected to get an expression like $n > \text{ some expression}$ so that I can plug that into 3. but it seems what I did in 4. is correct so I am not sure what I can do to get the $>$ into that expression.

EDIT:

$n\ge N(\epsilon,x)$ means that $n$ is greater than some number $N$ and that number $N$ is a function of $\epsilon$ and $x$.

Answer 1

I think you are going about this wrong. First, let $x \in (0,1)$, as $\ln(0)$ is not defined and $\ln(1)=0$, so you can't divide by $\ln(1)$. Next, let $\varepsilon>0$. There will exist an integer $n \in \Bbb{Z}$ satisfying $x^n < \varepsilon$. Now take the natural log of each side. You'll have $n\ln(x)<\ln(\varepsilon)$. By choice of $x$, $\ln(x)$ will be negative, so dividing by a negative flips the inequality around. The conclusion is that $n > \frac{\ln(\varepsilon)}{\ln(x)}$.

Now your numbers should make more sense. Except for where you chose $\varepsilon = 1$, $n=-1, x = 0.5$. This case is irrelevant, because you are only interested in $n$ that do satisfy the inequality, given $x$ and $\varepsilon$. Of course if you start with an $n$ that causes the inequality to fail, you will work through the math (as you did) to discover that the inequality fails!

As a final point, I have never seen the notation $n \geq N(\varepsilon,x)$. I won't say it's wrong because maybe it is used somewhere out there in the world, but it's odd to compare a single value to an entire interval with the $"\geq"$ notation. I'm assuming it means "$n$ is greater than or equal to every element in the interval $N(\varepsilon,x)$." I'd discourage writing a statement like that, and would instead write $n \geq x+\varepsilon$, as $x+\varepsilon$ is the smallest upper bound of the neighborhood $N(\varepsilon,x)$.