Continuity of the function $f(x)=\lim\limits_{n \to \infty}[\lim\limits_{t\to 0}[\frac{\sin^2 (n!\pi x)}{\sin^2(n! \pi x)+t^2}]] $, $x \in \mathbb R$

Question

Consider the Question. I think that f(x)=1 $\forall x \in \mathbb {R}$.

Reason: As t $\to$ 0 , $[\frac{\sin^2 (n!\pi x)}{\sin^2(n! \pi x)+t^2}]$ $\to[\frac{\sin^2 (n!\pi x)}{\sin^2(n! \pi x)}]= 1$.

So it is differentiable in $\mathbb R$. But the explanation given is the following.

Now my questions are as follows:

1) Am I right? if not, in what way?

2) Whether the explanation given in the picture is correct?

3)Can I interchange Limits? if So under what conditions.

Thanks in advance.

Answer 1

The expression you mention does not exists when $\sin\left(n! \pi x\right)=0$ so your thoughts are not bad, but are not true for all $x$. If $n!x \notin \mathbb{Z}$ then you are right $$ \frac{\sin^2\left(n! \pi x \right)}{\sin^2\left(n! \pi x \right)+t^2} \underset{t \rightarrow 0}{\sim}\frac{\sin^2\left(n! \pi x \right)}{\sin^2\left(n! \pi x \right)}=1 $$ However if $n!x \in \mathbb{Z}$, then $n!x=N_n$ then

$$\sin\left(n! \pi x\right)=\sin\left(N_n \pi \right)=0$$

Hence $f(x)=0$ if $n!x \in \mathbb{Z}$.

You can change limits when the sequence $$f_n(t)=\displaystyle \frac{\sin^2\left(n! \pi x \right)}{\sin^2\left(n! \pi x \right)+t^2}$$ converges uniformly on $I$ and if $t \in \overline{I}$.