Prove the $\lim_{n \to \infty} 2^\frac{1}{n} $ is $1$

Question

How do I prove the $\lim_{n \to \infty} 2^\frac{1}{n}$ is 1. I am only allowed to use the addition, division and multiplication property of limits, and the sqeeuze theorem. Specifically, the theorems I am allowed to use are

Define: $\lim_{n \to \infty} a_n = a$, $\lim_{n \to \infty} b_n = b$

1. $\lim_{n \to \infty} (a_n + b_n) = a + b$
2. $\lim_{n \to \infty} (a_n b_n) = ab$
3. $\lim_{n \to \infty} (\frac{a_n}{b_n}) = \frac{a}{b}$
4. if $\forall n\in \mathbb N: a_n \leq b_n \leq c_n $ and $\lim_{n \to \infty} a_n$ = $\lim_{n \to \infty} c_n $ = $C$, then $\lim_{n \to \infty} b_n = C$

With only these 4 theorems, I am quite stuck in thinking of how to apply them for limits involving exponents.

Answer 1

You can use the following elementary calculations:

$$1=2-1 = \left(2^{\frac 1n}\right)^n - 1= \left(2^{\frac 1n}-1\right)\sum_{k=0}^{n-1}2^{\frac kn}$$ Hence, $$0 \leq 2^{\frac 1n}-1 = \frac 1{\sum_{k=0}^{n-1}2^{\frac kn}} \leq \frac 1n$$

Answer 2

You should be aware of the episilon-N definition of infinite limits.

That is,$$\lim_{n\to \infty} x_n=A\iff \forall \epsilon>0\exists N\in\mathbb{N}\forall n>N(\mid x_n-A\mid<\epsilon) $$

Now, the trick of finding a limit is first use real values to find out where it tends to, then, you rigorously prove it by using the equivalent definition above on the RHS.

By pressing on a calculator, increasing the value of natural $n$, we should find that its absolute difference with $1$ should get smaller and smaller. That is, $\mid 2^{\frac{1}{n}} -1\mid $ should get closer to closer to 0, as n gets bigger and bigger

To put this formally, $\mid 2^{\frac{1}{n}} -1\mid\to 0 $ as $n\to\infty$

To give a formal proof of the limit, using the hint from @Sine of the time,

Taking the limit to both sides, by the comparison theorem you provided in your set of rules we get:

$$1\le\lim_{n\to \infty} 2^{\frac 1n}\le 1$$

Which is,

$$\lim_{n\to \infty} 2^{\frac 1n}= 1$$

Answer 3

Before trying to compute the limit, do this simple trick:

$$ \lim_{n \to \infty} f(n) = l \implies \lim_{n \to \infty} \ln(f(n)) = \ln(l) $$

So by taking the logarithm you get $\lim_{n \to \infty} \ln(2)/n = \ln(l)$ (being $l$ the result of the original limit). From here it is easy to see that $1/n$ goes to $0$ as $n$ tends to infinity and $\ln(2)$ stays constant. Using the second property you stated, it is trivial that $\ln(l) = 0$ and therefore $l = 1$.