Given the sequence of functions $$f_n : [0,1] \rightarrow \mathbb{R},f_n(x) =\left\{\begin{array}{ll} n \sin(n^2x), & 0 ≤ x ≤ \frac{\pi}{n^2},\\ 0, & \text{else.} \end{array}\right. $$
Examine for pointwise convergence.
I know that $f_n$ converges pointwise to $0$, but I don't understand why.
In my thinkings $\displaystyle\lim _{n\to \infty }f_n$ equals to a factor $n$ which rises continuously multiplied with a value between $[-1,1]$.
Can someone explain how to derive pointwise convergence in this example?
The claim is that $f_n \to 0$ point-wise. Let $x_0 \in [0,1]$. If $x_0 = 0$, then we're done. So, we can assume that $x_0 > 0$. Then, there exists an $N \in \mathbb{N}$ such that: $$\forall n \geq N: \frac{\pi}{n^2} < x_0$$
This implies that for $n \geq N$, $f_n(x_0) = 0$. But this obviously just means that we have point-wise convergence to $0$ for this sequence of functions so we're done.