Proof of $\lim_{n \to \infty} a^{1/n} = 1$ and $\lim_{n \to \infty}b^{f(n)} = b^{\lim_{n \to \infty} f(n)}$ where $a>1$

Question

As title says, I am not sure what would be the proof of $$\lim_{n \to \infty} a^{1/n} = 1$$ would be where $a>1$. Also, how do you prove that $$\lim_{n \to \infty}b^{f(n)} = b^{\lim_{n \to \infty} f(n)}$$ where $b>0$?

Answer 1

Your first limit obviously follows from the second limit, so we will limit ourselves to proving the second.

$$\lim_{n\to\infty}b^{f(n)} = \lim_{n\to\infty}e^{f(n)\log(b)}$$

$e^x $ is continuous. Switch the limit. Q.E.D.

Answer 2

Hints :

Use the fact that :

$$a^b = \exp^{b\cdot \ln{a}}$$

And the fact that exp is a continuous function

Answer 3

1) If you want to argue more formally. For $a>1$, we have $a^{1/n}>1$. Define $c_n = a^{1/n}-1>0$, then $a=(1+c_n)^n\ge1+nc_n$. Thus $$0<c_n= a^{1/n}-1\le \frac{a-1}{n}\to0$$

Hence $a^{1/n}-1\to 0$, i.e., $a^{1/n}\to 1$.

2) Try to prove that $b^x$ is continuous if you want I could give a sketch of the proof.

Answer 4

Using Cauchy's second theorem on limits
For $a>1\in\mathbb{R}$, and sequence $x_{n}=a, \lim_{n\to\infty}\frac{x_{n+1}}{x_n} =\frac{a}{a} =1$
$\therefore \lim_{n\to\infty}a^{1/n}=1$

PS: Cauchy's second theorem on limits says that for a sequence $\{a_n\}$ of positive values, if $$\lim_{n\rightarrow \infty} \frac{a_{n+1}}{a_n} = \ell,$$ then $$\lim_{n\rightarrow \infty} (a_n)^{1/n} = \ell.$$