Find $\lim_{n\to \infty}\sqrt[n]{a}$ where 0 < a

Question

Find $\lim_{n\to \infty}\sqrt[n]{a}$

Disclaimer: Since this is a sequence the fact that n approaches infinity is obvious, therefore I omitted it in the limit notation.

My idea is to analyze two cases: 1) $ 1 < a$ and 2) $ 0 < a < 1$.
(1) $1 < a$-n $$\sqrt[n]{a} = 1 + t_n$$ And $t_n >0 $Now, I use the binomial formula to get $$a >1+nt_n \Rightarrow t_n<\frac{a-1}{n} \iff 0 <\lim t_n<\lim \frac{a-1}{n} \Rightarrow \lim t_n =0$$ And since $\lim t_n = 0 $ $\lim_{n\to \infty}\sqrt[n]{a}= 1$
Then, I will use the same method to prove the second case.
Is it a correct way to do this? If so, is there an easier way?

Answer 1

For $0 < a < 1$, there is a interesting trick. Bernoulli's inequality does not seem directly applicable.

However, if you write $a=\dfrac1{1+b}$ where $b = \dfrac1{a}-1 \gt 0$, then $\dfrac1{a^n} =(1+b)^n \ge 1+nb \gt nb $, then $a^n \lt \dfrac1{nb} \to 0 $.

Answer 2

Let $$\:a^n=e^{\ln \left(a^n\right)}=e^{n\cdot \ln \left(a\right)}$$ Then $$\sqrt[n]{a} =a^{\frac{1}{n}}=e^{\frac{1}{n}\ln \left(a\right)}$$ Also $$u = \lim _{n\to \infty }\left(\frac{1}{n}\ln \left(a\right)\right) = 0 \implies \lim _{u\to \:0}\left(e^u\right)=1$$ Therefore $$\lim_{n\to \infty}\sqrt[n]{a} = 1$$