for a>0and n∈ℕ, define $a^{1/n}$ to be that unique positive number b such that $b^n$=a. Prove that for a>0,$lim_{n\rightarrow\infty}$ $a^{1/n}$=1.

Question

I know I need to consider both cases a $\leq$ 1 and a>1.

This is what I have so far:

a>1

|$a^{1/n}$ -1|< $\epsilon$

$a^{1/n}$-1<$\epsilon$

$a^{1/n}$<$\epsilon$+1

1/n ln(a) < ln($\epsilon$+1)

n>ln(a)/ln(1+$\epsilon$)

is this correct? and how do I show the other case a $\leq$ 1

When $a=1$, there is nothing to prove. When $a<1$, you have $|a^{1/n} - 1| = 1-a^{1/n}$.
You want $1-\varepsilon < a^{1/n}$. So that $\ln(1-\varepsilon) < \ln(a)/n$ So you just want $n > \ln(a)/\ln(1-\varepsilon)$ — aleph0, Feb 25 '19 at 16:42

score 1 · Answer 1 · answered Feb 25 '19 at 16:39

Let $a>0$,

Notice that $a^{1/n} = e^{\ln(a)/n}$. As $n\to+\infty$, $\dfrac{\ln(a)}{n}\to 0$ (notice that you don't have to check if $a\ge 1$ or $a<1$).

Therefore, $e^{\ln(a)/n}\to 1$ as $n\to\infty$ by continuity of the exponential.

So that $\lim\limits_{n\to+\infty}a^{1/n} = 1$

1 Answers1