Convergence and uniform convergence of $f_n(x)$ The given sequence of functions is defined as $f_n(x) = \frac{x^n}{n+x^n}$ for $x\ge 0$ and $n = 1,2,\ldots$; let $f = \begin{cases}
0 &:0\le x \le 1\\
1 &:x>1
\end{cases} $
My hypothesis is that $f_n\to f$ point-wise, but not uniformly (the function $f$ was not given, I have defined it). I have managed to show $f_n \to f$ point-wise on $[0,1]$ but am having difficulties proving convergence for $x>1$. My hunch is that the $N$ we choose must depend on $\epsilon$ and $x$ (which is why I also believe $f_n$ does not converge uniformly), but I can't seem to find a suitable $N$. Any hints would be appreciated.
 A: Point-wise convergence for $x>1$ follows from
$$\lim_{n\to+\infty} \frac{x^n}{n+x^n}=1,$$
since $x^n$ grows exponentially with base $x>1$, hence it dominates for large $n$:
$$\frac{n+1}{x^{n+1}}\Big/ \frac{n}{x^n}= \frac{n+1}{nx}<1$$
for any $n$ such that $n>1/(x-1)$. Let $n_0$ the minimum $n$ such that $n>1/(x-1)$.
Thus for any $n\geq n_0$ it holds
$$\frac{n+1}{x^{n+1}}\Big/ \frac{n}{x^n}= \frac{n+1}{nx}<1,$$
hence
$$\frac{n_0+m}{x^{n_0+m}}\Big/ \frac{n_0}{x^{n_0}}\leq \left(\frac{n_0+1}{n_0x}\right)^m.$$
Fix $\epsilon>0$: 
$$\frac{n_0+m}{x^{n_0+m}}\leq \frac{n_0}{x^{n_0}}\left(\frac{n_0+1}{n_0x}\right)^m<\epsilon$$
whenever $m>\log_{\frac{n_0+1}{n_0x}} (\epsilon \frac{x^{n_0}}{n_0})$.
You are also correct about non-uniform convergence: fix an arbitrary $\epsilon>0$, if uniform convergence holds, then there exists $N$ such that for any $n\geq N$, $\sup_x|f_n(x)-f(x)|<\epsilon$. However for any $n$ you have
$$\lim_{x\to 1^+} 1-\frac{x^n}{n+x^n} = \lim_{x\to 1^+} \frac{n}{n+x^n} =\frac{n}{n+1}.$$
A: For $x>1$, we have $\frac{x^n}{x^n+n}=\frac{1}{1+(\frac{n^{1/n}}{x})^n} \in \left[\frac{1}{1+(\frac{1+(x-1)/2}{x})^n},1\right] = \left[\frac{1}{1+a^n},1\right]$ for sufficiently large $n$, where $a=\frac{1+(x-1)/2}{x}<1$. Further, by Squeeze theorem, $\frac{x^n}{x^n+n}$ converges to 1.
Then let $(x_n)=(n^{1/n})>(0)$. But $f_n(x_n)=(\frac{1}{2})$ does not converge to 1.
In fact, if you have already learnt that "continuous functions which converge uniformly must converge (also pointwise) to a continuous function", then you can conclude immediately that the convergence was not uniform, because $f$ is not continuous.
