Pointwise prove. Prove that $$f_n =\begin{cases} n\sin(nx) &\text{for} \space 0 \leq nx \leq \pi\\ 0 & \text{otherwise}\end{cases}$$ converges pointwise to 0 as $n \to \infty$, being n a integer and for all x satisfying the properties.
I retired this question from a book about complex numbers/functions exercise.
Now i just get a counterpoint, see: Adote $x = \pi/2n$, so that $f_n = nsin(\pi/2)=n$. Now this certainly does not tends to 0. What am i doing wrong? I mean, i just found a counterprove to what i should prove, so i think or the enunciate is wrong or am i wrong, so, if i am wrong, where is my error?
 A: Your argument only proves that $f_n$ does not converge uniformly. But it does converge pointwise.
If $x >0$ then $0 \leq nx \leq \pi$ will never hold for large enough $n$: To be specific for $n >\frac {\pi} x$. So $f_n(x)=0$ for $n >\frac {\pi} x$ proving that $f_n(x) \to 0$. Also $f_n(x)=0$ for all $n$ if $x \leq 0$.
A: Pointwise convergence to $0$ is that for any fixed real number $x$, the sequence $|f_n(x)|$ converges to $0$. To negate this, you would have to locate some specific $a$ such that $|f_n(a)|$ does not converge to $0$. You have done something different, you have found a sequence $x_n = \frac{\pi}{2n}$ such that $|f_n(x_n)|$ does not converge to $0$. The key difference is that $x_n$ is varying as $n$ changes. Pointwise convergence is strictly about each individual fixed point, but you have a point which varies.
But don't discount what you have done! The idea of a sequence of functions converging is much deeper than pointwise convergence. Asking how the convergence relates between different points is a very good thing to do. In particular, what you have shown is that the sequence of functions $f_n$ does not converge uniformly to $0$. This is due to a general fact which says that if $f_n \to f$ uniformly and $x_n \to x$ then $f_n(x_n) \to f(x)$. This is the fact you contradicted, so you have shown that this sequence of functions is not uniformly convergent. If you're not familiar with this notion, I'd encourage you to look into it, as you've now come across a key idea.
A: What you're doing wrong is that to evaluate pointwise convergence, you have to fix $x$ and then look at the convergence of the sequence $\{f_n(x)\}$. $x= \pi/2n$ depends on $n$.
Here, you can proceed as follows.
For $x \le 0$, $\{f_n(x)\}$ is the always vanishing sequence. So the conclusion is immediate.
And for $x \gt 0$ notice that $\{f_n(x)\}$ is an eventually vanishing sequence that vanishes for $ n \gt \lfloor \frac{\pi}{x} \rfloor$.
In both cases, we get the conclusion that $f_n(x) \to 0$ as desired.
