Does $\frac{nx}{1+n \sin(x)}$ converge uniformly on $[a,\pi/2]$ for all $a \in (0,\pi/2]$?

Question

Edit: the question had some missing details. It should read as follows:

Prove for all $a \in (0,\frac{\pi}{2}]$, $f_n \rightarrow f$ uniformly on $[a,\frac{\pi}{2}]$. Here $$f_n(x) = \frac{n x}{1 + n \sin(x)}.$$

Here is my attempt at the problem:

If $x \in (0, \frac{\pi}{2}]$ then $| f_n(x) - f(x) |$

$$ =\left| \frac{nx}{1 + n\sin(x)} - \frac{x}{\sin(x)}\right| $$

$$ = \left|\frac{nx \sin(x) - x(1 + n \sin(x))}{[1+ n \sin(x)]\sin(x)}\right|$$

$$ = \frac{x}{\sin(x) + n \sin^2(x)} \leq \frac{1}{n} \space \space \text{(is this line correct)}$$

So $\forall \epsilon > 0$, we may choose $N \geq \frac{1}{\epsilon}$ such that when $n \geq N \implies |f_n(x)-f(x)| \leq \epsilon \space \space \forall x \in (0, \frac{\pi}{2}]$

It's already been established that

$f_n(x)$ converges pointwise to: $f(x) = 0, x =0$ and $f(x) = \frac{x}{\sin(x)} ,x \in (0, \frac{\pi}{2}]$

Answer 1

One has $${\rm sinc}(x)-f_n(x)={1\over n}{x\over \sin x\bigl(\sin x+{1\over n}\bigr)}\ .$$ Now if an $a>0$ is given then $$0\leq{x\over \sin x\bigl(\sin x+{1\over n}\bigr)}\leq{\pi/2\over\sin^2 a}=:C\qquad\bigl(a\leq x\leq{\pi\over2}\bigr)\ ,$$ and therefore $$\bigl|{\rm sinc}(x)-f_n(x)\bigr|\leq{C\over n}\qquad\bigl(a\leq x\leq{\pi\over2}\bigr)\ .$$

Answer 2

If the convergence were uniform, then as $n\to \infty,$

$$\sup_{x\in(0\pi/2]}\,|f_n(x) - x/(\sin x)| \to 0\implies |f_n(1/n) - (1/n)/(\sin (1/n))| \to 0.$$

But $f_n(1/n) \to 1/2$ and $(1/n)/(\sin (1/n)) \to 1.$

Answer 3

We have that $$\forall x>0\:\:\forall \varepsilon>0\:\:\exists N_{\varepsilon,x}>0\:\:\forall n\ge N_{\varepsilon,x}\implies\left|\frac{nx}{1+n\sin (x)}-\frac{nx}{1+nx}\right|<\varepsilon\:\frac{n|x|}{|1+nx|}\\\iff|x-\sin(x)|<\varepsilon\left|\frac{1}{n}+\sin(x)\right|\\\iff\frac{nx}{1+n\sin (x)}\underset {\:\:0^+}\sim\frac{nx}{1+nx}.$$

Now we have to guess the pointwise limit and prove that it holds in a rigorous fashion : $$\forall x>0\:\:\forall \epsilon>0\:\:\exists K_{\epsilon,x}>0\:\:\forall n\ge K_{\epsilon,x}\implies\left|\frac{nx}{1+nx}-1\right|<\epsilon,\\\text{as long as}\:\:\:K_{\epsilon,x}>\frac{1}{|x|\epsilon}=\frac{1}{x\epsilon}.$$

Maybe there's some problem with uniform convergence as $\:x\to0^+.$

Our best bet is to choose some point that'll try to reach the lower bound in the interval : $$\:x_n=\frac{1}{n}\in\:\left(0,\frac{\pi}{2}\right]\overset{\text{let}}=E\\\implies\frac{nx_n}{1+nx_n}=\frac{1}{2}.$$

To be formal, $$\exists\hat\varepsilon>0\:\:\forall M_{\hat\varepsilon}>0\:\:\exists x_n\in E\:\:\exists n\ge M_{\hat\varepsilon}\implies|f_n(x_n)-f(x_n)|\ge\hat\varepsilon.$$