I need help in this one.

Let $f\_n(x)=x^{1/n}$ for $x\in[0,1]$. Then which one of the following is true?

$1.$ $\lim\_{n\to\infty}f(x)$ exists for all $x\in[0,1]$.

$2.$ $\lim\_{n\to\infty}f(x)$ defines a continuous function on $[0,1]$.

$3.$ ${f\_n}$ converges uniformly on $[0,1]$.

$4.$ $\lim\_{n\to\infty}f(x)=0$ for all $x\in[0,1]$.

My attempt: f$(0)=0$ and $f(1)=1$, it would be so even for large $n$. So, 4th option is out. And now with calculator, I can see that $0.001^{0.001}=0.99$ and $0.9^{0.001}=0.99$, so I am tended to assume that the limit gives us a function which is $0$ at $x=0$ and $1$ at other values in the domain. Clearly, it's not continuous at $x=0$, so $2$nd and $3$rd options are also ruled out. So answer is $1$st option.

Is my answer correct? And can anyone please suggest me a way which is free of calculator?