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.

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 $\displaystyle\lim_{n\to\infty}\frac{\ln 2}{n}$ ?
