Uniform convergence of $f_n(x)=\frac{1}{1+nx}$ on $(0,1)$ Consider the sequence of functions $f_n(x)=\frac{1}{1+nx}$ for $x\in (0,1)$. Then 


*

*$f_n (x) → 0$ pointwise but not uniformly on $(0,1)$.

*$f_n (x) → 0$ uniformly on $(0,1)$.

*$\int_{0}^{1} f_n (x) dx → 0$ as $n → ∞$.

*$f'_n(1/n)\to 0$ as $n\to \infty$.


I have shown that $3$ and $4$ are false. As $\lim_{n\to \infty} \int_{0}^{1}f_n(x)dx=1$ and $f'_n(1/n)=-\frac{n}{4}$. But I am confused about the convergence part. I know $f_n(x)$ is not uniformly convergent on $[0,1]$, so can I just remove the end points and claim that it is also not uniformly convergent on $(0,1)$ as well. It does not feels rigorous. How can I do this. Please help me out. 

Let $\epsilon=\frac{1}{2}$, $n_k = k+1$, and $x_k=\frac{1}{k+1}$, then
\begin{align}
|f_{n_k}(x_k)-f(x_k)| &= \left|\frac{1}{1+(k+1)\cdot\frac{1}{k+1}}\right|\\
&=\frac{1}{2}\\
&\ge \epsilon,
\end{align}
so $(f_n)$ does not uniformly converge.
