I am fining the pointwise limit of the function $f_n(x) = \frac{x^n}{3-x^n}$ for $x ∈ [0,1]$ and $n ∈ N$
In order to do this I first divided through by $x^n$, yielding me $$ \frac{1}{\frac{3}{x^n}-1}$$
Using this I have determined that $f_n(x) \to f(x) := \begin{cases} 0,\ 0 \leq x<1 \\ \frac{1}{2},\ x=1 \end{cases}$
If I assume that I have done this correctly then I could deduce that the convergence cannot be uniform on $[0, 1]$ since each $f_n$ is continuous, but the limit function is not continuous.
Am I correct to make this assumption or is there actually uniform convergence? If there is, how is it determined?
Due to this I have also adjusted the bounds of $x$ to $[0,\frac{1}{2}]$ to see if this instead would have uniform convergence.
In this case we would have $f_n(x) \to f(x) := \begin{cases} 0,\ 0 \leq x≤\frac{1}{2} \end{cases}$
What are the implications for the uniform convergence of this? Surely this is also not uniform convergence for similar reasons as above?
I am struggling to get my head around all this so any help would be greatly appreciated!
Uniform convergence of $(f_n)_{n\in \Bbb N}$ to $f$ on $[0,1]$ means $$\lim_{n\to \infty}\|f-f_n\|=0$$ where $\|f-f_n\|=\sup\{|f(x)-f_n(x)|: x\in [0,1]\}.$
Each $f_n$ is continuous with $f_n(0)=0$ and with $f_n(1)=1/2$ so there exists $x_n\in (0,1)$ with $f_n(x_n)=1/4.$
But $f(x_n)=0$ so $$\|f-f_n\|\ge |f(x_n)-f_n(x_n)|=|0-1/4|=1/4.$$ So $(\|f-f_n\|)_{n\in \Bbb N}$ does not converge to $0.$
Your reasoning, that if the convergence were uniform then $f$ would be continuous, is correct. However, in some cases, such as in this problem, we can bypass that.