# Convergence of $f_n(x) = \frac{nx}{1+n^2x^2}$

(Disclaimer: This is a homework question.) $$\text{For }n\text{ in }\mathbb{N}\text{ let }f_{n}:\mathbb{R} \rightarrow \mathbb{R} \text{ be defined by } f_{n}(x)=\frac{nx}{1+n^2x^2}$$

Show that the sequence $(f_{n})_{n\in\mathbb{N}}$ is convergent. Is this convergence uniform?

My attempt:
as $n \rightarrow \infty\quad f_{n} \rightarrow \frac{1}{nx} \implies f_{n} \rightarrow f = \left\{\begin{matrix} &0 &x\neq0 \\ &?? &x = 0 \end{matrix}\right.$
It seems to me $f$ is discontinous at $x=0$, unless I misunderstood.
For uniform convergence it's required that:$$\forall \varepsilon>0\quad \exists K_{\varepsilon}\in\mathbb{N} \quad s.t.\quad \forall n\geq K_{\varepsilon}\\||f_{n}(x)-f(x)||<\varepsilon$$ if we pick $n=k$ and $x_{k} = \frac{1}{k}$ then $$f_{n} = 1\quad \forall n\in\mathbb{N}\\\implies||f_{n}(x)-f(x)|| = ||1-0||=1$$ Does this imply $f_{n}$ is not uniformly convergent?

• What is the domain of this sequence of functions? Commented Dec 12, 2017 at 1:07
• you have $f_n(0)=0$ for all $n$ so $f(0)=0$.
– zwim
Commented Dec 12, 2017 at 1:11
• Right, your limit is funky. You wrote as $n \to \infty$ $f_{n} \to \frac{1}{nx}$, but if you had to actually do this limit in say a calculus course, would you not write that $f_{n} \to 0$? Commented Dec 12, 2017 at 1:12