Question: I have been trying to solve this problem for hours...  

At this point, I believe that there is a mistake in the problem, but I assume that I am in the one who is wrong. May you help me?

***Here is the problem:***

Let $f\_n(x)= \frac{nx}{3+nx},\ n\in \mathbb{N},\ \text{and}\ x>0.$

***a) Show that the sequence $\{f\_n\}\_{n\in \mathbb{N}}$ converges uniformly on $[a, \infty)$ for any $a>0$***

***b) Show that the sequence $\{f\_n\}\_{n\in \mathbb{N}}$ does not converge uniformly on $\mathbb{R}$.***

Here is my perspective:

To prove that the sequence converges uniformly I need to find an $N$ that depends only on epsilon. I am using this definition:

[![Uniform Convergence definition](https://i.stack.imgur.com/DRnzb.jpg)](https://i.stack.imgur.com/DRnzb.jpg)

I made a graph with some of the functions:

Here is the graph:

[![Graph of functions](https://i.stack.imgur.com/DFuGq.jpg)](https://i.stack.imgur.com/DFuGq.jpg)

According to the graph, from $[1, \infty)$, the limit of the $f\_n$ is $1$, but when $x$ is close to 0 the limit, from the right, is $0$.

Then I am wondering is the a of the problem $[a, \infty)$ must be greater or equal than 1 instead of greater than 0 to get uniform convergence.

When I calculate absolute value of $f\_n(x) - f(x),$ 
I am getting $nx > \frac{3}{\varepsilon} - 3$.

I do not see how to find an N that depends only on epsilon.

May you tell me if my reasoning is correct?

Answer: 
You are absolutely correct that the problem is at $x=0.$ However, you're asked to establish uniform convergence on $[a,\infty)$ for $a>0$, so the problem at zero isn't a problem on this interval. Note that for any $a>0,$ $f\_n(a)\to 1$ so it's not the case that points close to $x=0$ have $f(x)\to 0$; this only happens at $x=0$ itself. The points closer to zero will go more slowly, though.

As a reflection of this, notice from your plots that convergence to $1$ improves as $x$ gets larger. This actually helps you see that there's uniform convergence. The $N$ that works for $x=a$ will also work for $x>a.$ Now, you just need to convert this into a rigorous proof.

And then to show that it doesn't converge uniformly on $[0,\infty),$ just use the problem you noticed at $x=0.$ The function that you converge to pointwise isn't continuous, and this always means that the convergence isn't uniform (if each $f\_n$ is continuous).
For all $n\in\mathbb{N}$, let $$f\_n:(0,+\infty)\to\mathbb{R},x\mapsto\frac{nx}{3+nx}$$
The sequence $(f\_n)$ is pointwise convergent to $L:x\mapsto 1$.

To decide whether the convergence is uniform or not, consider :

$$\delta\_n(x)=\left|L(x)-f\_n(x)\right|=1-\frac{nx}{3+nx}=\frac{3}{3+nx}$$

We see that $\delta\_n\left(\frac 1n\right)=\frac 34$, hence $$\sup\_{x>0}\delta\_n(x)\not\to 0$$

and the convergence is not uniform on $(0,+\infty)$.

However, if we fix $a>0$ and consider $x\in[a,+\infty)$ :

$$\delta\_n(x)\le\frac{3}{3+na}$$

Hence :

$$\sup\_{x\ge a}\delta\_n(x)\le\frac{3}{3+na}\to 0$$

which proves that the convergence is uniform on any $[a,+\infty)$, with $a>0$.