I have been buggin on this question for a few hours now.
Q. for n = 1, 2, 3, ... let $f_n : [0, +\infty) \rightarrow \mathbb R$ be defined by
$f_n = \frac{xn + sin (\frac{x}{n^2})}{2(n + \sqrt x )}$
Show that $f_n$ converges pointwise on $[0, +\infty)$ and determine whether $f_n$ converges uniformly on $[0, +\infty)$ or not.
I have tried to come up with a solution on my own but I'm not sure if it's correct.
My attempt:
$\forall x \in [0, +\infty),$
$$\lim_{n \to \infty} f_n (x) = \lim_{n \to \infty} \frac{xn}{2(n + \sqrt x)} + \lim_{n \to \infty} \frac{sin (\frac{x}{n^2})}{2(n + \sqrt x)}$$
$$= \lim_{n \to \infty} \frac{xn}{n(2+\frac{2\sqrt x}{n})} + \lim_{n \to \infty} \frac{sin (\frac{x}{n^2})}{2(n+ \sqrt x)}$$
$$= \lim_{n \to \infty} \frac{x}{(2+\frac{2\sqrt x}{n})} + \lim_{n \to \infty} \frac{sin (\frac{x}{n^2})}{2(n+ \sqrt x)}$$
As $n \to \infty$, $\lim_{n \to \infty} \frac{x}{(2+\frac{2\sqrt x}{n})}$ becomes $\frac{x}{2}$ and $\lim_{n \to \infty} \frac{sin (\frac{x}{n^2})}{2(n+ \sqrt x)}$ becomes zero.
So $f_n \to f$ defined by $f : [0, +\infty) \to \mathbb R,$
$f(x) = \begin{cases} 0, & \text{if $x = 0$ } \\[2ex] \frac{x}{2}, & \text{if $x \neq 0$} \end{cases}$
So $f_n$ converges pointwisely.
However, for uniform convergence,
$$|f_n(x) - f(x)| = |\frac{xn + sin(\frac{x}{n^2})}{2(n + \sqrt x)} - \frac{x(n + \sqrt x)}{2(n + \sqrt x)}|$$
$$= |\frac{sin(\frac{x}{n^2})-x^\frac{3}{2}}{2(n+ \sqrt x)}|$$
if we set $x = n^2$
$$= |\frac{sin 1 - n^3}{2(n + \sqrt n^2)}| = |\frac{sin 1}{4n} - \frac{n^2}{4}|$$
$$\Rightarrow \lim_{n \to \infty} ||f_n - f|| \neq 0$$
Hence $f_n$ does not converge uniformly.
$ $
Is my solution reasonably correct?
Thanks