Is this epsilon delta argument that $\lim_{n \to\infty} (0.3)^{\frac{1}{n}} = 1$ valid?

Question

I'd like to know if this proof that $\lim_{n \to \infty} (0.3)^{\frac{1}{n}} = 1$ is valid. My approach seems too easy.

For $\epsilon >0$, I need to find an $n_0$ such that whenever $n>n_0$, we have $\lvert (0.3)^\frac {1}{n}-1\lvert<\epsilon$. So trying to simplify, I got $$-1<(0,3)^\frac {1}{n}<\epsilon+1$$ Therefore, $$(0.3)<(\epsilon+1)^{n}$$ $$(0.3)<1+n\epsilon \leq (1+\epsilon)^{n}$$ Since epsilon is greater than $0$ we know that $(0.3)<1+n\epsilon$ is true for all $n>0$. Therefore, given $\epsilon >0$, let $n_0>0$ and choose $n>n_0$.

Edit: I used Bernoulli's inequality to introduce this result $1+n\epsilon \leq (1+\epsilon)^{n}$

Answer 1

Here is a simpler argument, not original with me, that works for any $0 < a < 1$. Your case is $a = .3$.

Since $0 < a < 1$, there is a $b > 0$ such that $a = \frac1{1+b}$. Explicitly, $b = \frac1{a}-1$.

Then, by Bernoulli, $(1+b)^n \ge 1+bn \gt bn =n(\frac1{a}-1) $ so $a^n =\dfrac1{(1+b)^n} \lt \dfrac1{n(\frac1{a}-1)} = \dfrac{a}{n(1-a)} $.

Answer 2

Note that $$|(0.3)^{1/n}−1|<ϵ \iff 1-\epsilon <(0.3)^{1/n} <1+ϵ $$

Apparently you have missed some parts in this inequality which made your argument questionable.