Checking uniform convergence in a proper way

Question

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?

Answer 1

Yes your answer is correct but you should add that for $0<x\leq 1$ we have $$\lim_{n\to\infty}x^{1/n}=\lim_{n\to\infty}e^{\frac{1}{n}\log x}=1$$ so the limit function $f$ is defined by $$f(x)=\left\{ \begin{array}\\ 1&\text{if}\ 0<x\leq 1\\ 0&\text{if}\ x=0 \end{array}\right.$$ and since $f$ isn't continuous on $[0,1]$ the convergence isn't uniforme.

Answer 2

Clearly $f_n(0)=0$ for all $n$. Thus $\lim_{n\to \infty} f_n(0)=0$. So what happens when $x\neq 0$?

We use that $x^y$ is a continuous function of $y$ if $x$ is a fixed number. By a standard theorem on limits we get $\lim_{n\to \infty} x^{\frac 1 n } = x^0=1$.

Accordingly, option 1 is true.