Function how to show whether it is pointwise or uniform convergent?

Question

My sequence is : $s_n=\frac{2}{3+n|x|}$ for real x and n $\in \mathbb{N}$

Now I have done this:

$\lim_\limits{n \to 0} \frac{2}{3+n|x|}= \frac{2}{3}$

$\lim_\limits{ n \to \infty} \frac{2}{3+n|x|}= 0$

Now it is pointwise convergent to these 2.

Now I want to show if it uniform convergent or not:

$\left|\frac{2}{3+n|x|}-0 \right| \leqslant \left|\frac{2}{nx}\right| \leqslant \left|\frac{2x}{nx}\right|\leqslant \left|\frac{2}{n}\right| $

Now it is with this way because I have found a function for which this applies.

Is this correct? Can I simply multiply the upper part by $2$ to make it independent from $x$ ?

Edit: Now I have this

For x= 1/n

$\lim_\limits{ n \to \infty} sup \left|\frac{2}{3+n|1/n|}-0 \right| = 2/3 \neq 0 \ $

So it is not uniform convergent.

Now my second function

The sequence is : $s_n=\frac{2|x|}{3+n|x|}$ for real x and n $\in \mathbb{N}$

I have done this:

$\lim_\limits{n \to 0} \frac{2x}{3+n|x|}= \frac{2|x|}{3}$ Is this now still pointwise convergent due to the function?

$\lim_\limits{ n \to \infty} \frac{2|x|}{3+n|x|}= 0 ?$

$\lim_\limits{ n \to \infty} sup \left|\frac{2|x|}{3+n|x|}-0 \right| \leqslant \left|\frac{2|x|}{nx}\right| \leqslant \left|\frac{2}{n}\right| $

So this is now 100% uniform convergent, right?

Answer 1

Most easy is to use sufficient and necessary condition of uniform convergence $$\lim_\limits{n \to \infty}\sup_{x \in \mathbb{R}}|f_n(x)-f(x)|=0$$ In your example function $s_n(x)=\frac{2}{3+n|x|}$ is even on $\mathbb{R}$, and you can find, that sup reached in $x=0$. Hence?