I have a sequence of functions defined as $f_n:\mathbb R \rightarrow \mathbb R$ with $f_n(x) = \frac{nx}{1+\lvert{nx}\rvert}$
I am supposed to show that this does converge pointwise, but not uniformly. I have never really done this before and am kinda struggling with these exercises. I have seen some graphs from different sequences of functions and intuitively it does make sence. I just struggle computing it correctly.
So I recall having to calculate the pointwise limit function $f$. In other words, showing that $\lim_{n\to\infty}f_n(x) = f(x), \space \forall x \in \mathbb R$
It seems that I have to check for different cases here. My guess is $\lvert{x}\rvert \lt 0$, $\lvert{x}\rvert = 0$ and $\lvert{x}\rvert >0$
Showing that a sequence of functions converges pointwise, we have to show that $\forall x \in \mathbb R: \forall \varepsilon \in \mathbb R_{>0}: \exists N \in \mathbb R_{>0}: \forall n > N: \lvert{{f_n}(x)-f(x)}\rvert < \varepsilon$.
To calculate $f(x)$, we have to first assume a limit I believe. Since we have three different possible cases, we have to differentiate between them.
So take $x=0$, we get $\frac{n*0}{1+\lvert{n*0}\rvert} = \frac{0}{1} = 0$. Am I making a crucial mistake here already or is this correct and I can proceed from this? I'm not really sure since now $f(x) = 0$ and therefore $\lvert{{f_n}(x)-f(x)}\rvert = \lvert{{f_n}(x)-0}\rvert$ = $\lvert{{f_n}(x)}\rvert$
For $\lvert{x}\rvert \lt 0$ and $\lvert{x}\rvert \gt 0$, my guess is that $f(x) = -1$ and $1$. However, again, I'm really unsure about my calculations and it seems that I did not quite understand this enough. Since this is where I'm stuck and do not know how to proceed.
My questions now are: is my work partially correct or did I make some crucial mistakes early on? And also, how can I proceed from this or if it's wrong, how can I solve an exercise like this?