Limit of $2^{1/n}$ as $n\to\infty$ is 1 How do I prove that: 
$\lim \limits_{n\to \infty}2^{1/n}=1$
Thank you very much.
 A: Note that for $a\gt 0$, 
$$a^b = e^{b\ln a}.$$
So
$$\lim_{n\to\infty}2^{1/n} = \lim_{n\to\infty}e^{\frac{1}{n}\ln 2}.$$
Since the exponential is continuous, we have
$$\lim_{n\to\infty}e^{\frac{1}{n}\ln 2} = e^{\lim\limits_{n\to\infty}\frac{1}{n}\ln 2}.$$
Can you compute $\displaystyle\lim_{n\to\infty}\frac{\ln 2}{n}$ ?
A: Using the binomial theorem for integer exponents:
Can you see that $(1+\frac 1 n)^n > 2>1$
Take the nth root, to give:
$(1+\frac 1 n) > 2^{\frac 1 n} >1$
A: HINT:
Use squeeze theorem.
Since $1 < 2$, we have $1 = 1^{1/n} < 2^{1/n}$ for all $n \in \mathbb{N}$.
To bound the limit from above, note that $1 + n \epsilon < \left( 1 + \epsilon \right)^n$.
Hence, given any $\epsilon >0$, $\forall n > \displaystyle 1/{\epsilon}$, we have $2 < 1 + n \epsilon < \left(1 + \epsilon \right)^n$ and hence $2^{1/n} < 1 + \epsilon$.
A: Here are a few different proofs which don't use $e$ or $\log$ and can be regarded completely elementary.
Proof 1)
We use the following theorem:
If $a_n \gt 0$ and $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L$, then $\lim_{n\to\infty} a_n^{1/n} = L$
This is a standard theorem, and a proof can be found in almost any textbook. You can also find a proof in my answer here: Show that this limit is equal to $\liminf  a_{n}^{1/n}$ for positive terms.
Apply the theorem to the sequence $a_n = 2$.
Proof 2)
We use the $\text{AM} \ge \text{GM}$ inequality on $n-1$ ones and one $2$.
$$\frac{1 + 1 + \dots + 1 + 2}{n} \ge 2^{1/n}$$
$$ \frac{n+1}{n} \ge 2^{1/n}$$
Thus we have
$$ 1 + \frac{1}{n} \ge 2^{1/n} \ge 1$$
so by Squeeze theorem, $\lim_{n \to \infty} 2^{1/n} = 1$.
Proof 3)
We can use Bernouli's inequality (essentially similar to Sivaram's answer) to show that
$$\left(1 + \frac{1}{n}\right)^n \ge 1 + \frac{1}{n} \times n =  2$$
and we get inequalities similar to the proof in 2).
Proof 4)
The sequence $a_n = 2^{1/n}$ is bounded below (by $1$) and montonically decreasing.
Thus it is convergent, to say $L$. Since $a_{2n}$ also converges to $L$, we have that $L = \sqrt{L}$, as $2^{1/2n} = \sqrt{2^{1/n}}$. So $L = 0$ or $L = 1$. Since the limit is not less than $1$ ($2^{1/n} \ge 1$), the limit is $1$.
Proof 5)
For $n \gt 2$, we have that $1 \le 2^{1/n} \le n^{1/n}$.
Now use the fact that $\lim_{n \to \infty} n^{1/n} = 1$.
An elementary proof of that can be found here: https://math.stackexchange.com/a/115825/1102. Any proof for $n^{1/n}$ now becomes a proof for $2^{1/n}$. Proof 1) above can also be used for $n^{1/n}$.
Proof 6)
Using combinatorics.
The number of $n$ digit numbers in base-$n+1$ is $(n+1)^n$ (allowing for leading zeroes). The number of $n$ digits numbers in base-$n$ is $n^n$. We can show that $(n+1)^n \ge 2 \times n^n$: consider the base-$n$ numbers. Replace the last digit with $n$. You get a base-$n+1$ $n$ digit number. Counting the base-$n$ numbers (which are also base-$n+1$ numbers) and the "last digit modified" numbers, gives us the inequality.
This inequality implies that $1 + \frac{1}{n} \ge 2^{1/n}$ and can be used to give a proof using the squeeze theorem, similar to proofs 2 and 3.
