How to find the limit for $f_n$ within an interval $x \in [0,1]$ and whether the function converges uniformly in an interval $(0,1)$

Question

If there is a function $\frac{x^n}{1+x+x^2},$ where $x\in [0,1]$, then find $\lim_{n\to \infty}f_n$. Ok for $x=1$ $\lim_{n\to \infty} \frac{1^n}{3}=\frac{1}{3}$. For $x=0$: $\lim_{n\to \infty} \frac{0}{1} = 0$. But how do I proceed with the rest of the interval?

Answer 1

Point-wise convergence: $ 0 \leq \frac {x^{n}} {1+x+x^{2}} \leq x^{n}$ and $x^{n} \to 0$ for $0\leq x <1$.

The convergence is not uniform. Let us prove this by contradiction; suppose the convergence is uniform. Choose $0<\epsilon <\frac 1 {3e}$. Then there exists $n_0$ such that $\frac {x^{n}} {1+x+x^{2}} <\epsilon$ for all $x$ whenever $n >n_0$. Put $x=1-\frac 1 n$ to get $\frac {(1-\frac 1 n) ^{n}} {1+(1-\frac 1 n)+(1-\frac 1 n)^{2}} <\epsilon$ for all $x$ whenever $n >n_0$. Letting $n \to \infty$ in this we get $\frac 1 {3e} \leq \epsilon$ which is a contradiction.