Find the limit $ \lim_{n\to \infty} 2^{\frac{1}{n}}$ 
$$\lim_{n\to \infty}2^{\frac{1}{n}}$$

Basically looking at $$\lim_{n\to \infty}2^{\frac{1}{n}}=2^{\lim_{n\to \infty}{\frac{1}{n}}}=2^0=1$$
But if I use $$\lim_{n\to \infty}2^{\frac{1}{n}}=\lim_{n\to \infty}e^{\ln\left(2^{\frac{1}{n}}\right)}=\lim_{n\to \infty}e^{{\frac{1}{n}}\cdot{\ln(2)}}=\lim_{n\to \infty}e^{\frac{1}{n}}\cdot\lim_{n\to \infty}e^{\ln(2)}=1\cdot2=2$$
Where did I got it wrong?
 A: Rules of exponentiation:


*

*$a^{b+c}=(a^b)\times(a^c)$

*$a^{b\times c}=(a^b)^c$



Therefore:


*

*$\lim\limits_{n\to\infty}e^{{\frac{1}{n}}+{\ln2}}=\left(\lim\limits_{n\to\infty}e^{\frac{1}{n}}\right)\times\left(\lim\limits_{n\to\infty}e^{\ln2}\right)$

*$\lim\limits_{n\to\infty}e^{{\frac{1}{n}}\times{\ln2}}=\left(\lim\limits_{n\to\infty}e^{\frac{1}{n}}\right)^{\lim\limits_{n\to\infty}\ln2}$

A: $$\lim_{n\to \infty} e^{\frac{1}{n}\cdot \ln 2} \neq \displaystyle \lim_{n \to \infty} e^{\frac{1}{n}}\cdot \displaystyle \lim_{n \to \infty} e^{\ln2}$$
A: Personally I prefer to do it this way
$$\lim_{n\to \infty} e^{\frac{1}{n}\cdot\ln2}=\lim_{n\to \infty}\bigl( e^{\ln2}\bigr)^{\frac{1}{n}}=\bigl(e^{\ln2}\bigr)^{\lim_{n\to \infty}\frac{1}{n}}=\bigl(e^{\ln2}\bigr)^0=1$$
But maybe thats just me. the other ways work too
A: See $\lim \limits_{n - \infty}e^{\frac{1}{n}*\ln2}=\lim \limits_{n - \infty}(e^{\frac{1}{n}})^{\ln2}=1^{\ln2}=1$
A: Let $\alpha_n$ be a convergent sequence with non-negative terms  such that $$2^{1/n}=1+\alpha_n.$$
Then, we have, $$2=(1+\alpha_n)^n=1+n\alpha_n+\frac {n(n-1)}2\alpha_n^2+\dots\\\implies2\gt1+n\alpha_n\\\implies\alpha_n\lt \frac 1n$$
So, we have $$0\lt \alpha_n\lt \frac 1n.$$
So, $$\alpha_n\to 0.$$
So, $$\lim 2^{1/n}=\lim(1+\alpha_n)=1+\lim \alpha_n=1+0\\\implies \lim2^{1/n}=1$$
