### CSIR JUNE 2011 PART B QUESTION 24 SOLUTION (Convergence of $f_n(x) = x^{\frac{1}{n}}$)

Let $f_n(x) = x^{\frac{1}{n}}$ for $x \in [0,1]$. Then
1)$\lim\limits_{n \to \infty}f_n(x)$ exists for all $x \in [0,1]$,
2)$\lim\limits_{n \to \infty}f_n(x)$ defines a continuous function on $[0,1]$,
3)$f_n$ converges uniformly on $[0,1]$,
4)$\lim\limits_{n \to \infty}f_n(x) = 0$ for all $x \in [0,1]$.
Solution: Let $f(x) = \lim\limits_{n \to \infty}f_n(x)$. Let $x \in [0,1]$.  If $x=0$ then the sequence $\{x^{\frac{1}{n}}\}$ converges to $0$ and so $f(0)=0$. If $x=1$ then the sequence $\{x^{\frac{1}{n}}\}$ converges to $1$ and so f(1)=1 (so option (4) is false). In general, if $a>0$ then $a^{\frac{1}{n}}$ converges $1$. So the sequence $\{x^{\frac{1}{n}}\}$ converges to $1$ for any $x \in (0,1)$. This shows that the pointwise limit function $f(x) = \begin{cases} 0 \text{ if } x=0 \\ 1\, \text{ if} x \in (0,1] \end{cases}$. Hence the given sequence converges pointwise to $f$. Also, we note that $f$ is not a continuous function on $[0,1]$ and option (2) is false.
Each $f_n$ is continuous, but the limit function $f$ is not continuous.
Result: If $\{f_n\}$ is a sequence of continuous functions converges uniformly to a function $f$ then the function $f$ is continuous.
In our case, $f$ is not continuous and hence the convergence is not uniform and option (3) is false.
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