How to prove that this limit is equal to $0$? (I am working with $\mathbb{K}=\mathbb{R})$ Let $f\in \mathcal{C}[0,1]$ (being this set formed by all $f:[0,1]\to \mathbb{K}$ that are continuous); for each $n\in\mathbb{N}$, we define $g_{n}(t):=f(\frac{t}{n})\sin^{n}t$, for all $t\in[0,1]$. Prove that, in $H$, $\lim_{n\to\infty}g_{n}=0$.
My attempt: I have tried to prove that $||g_{n}||\to 0$, or equivalently, that $||g_{n}||^{2}\to 0$; taking on account that that norm is the induced norm by the dot product in this preHilbert space (that is defined as: $\langle f,g\rangle=\int_{0}^{1} f(t)\overline{g(t)}dt$ , but in this case, with our choice of $\mathbb{K}$, is just $\langle f,g\rangle=\int_{0}^{1} f(t)g(t)dt$ ), $||g_{n}||^{2}=\int_{0}^{1} (g_{n}(t))^{2}dt=\int_{0}^{1} (f(\frac{t}{n}))^{2}\cdot \sin^{2n}(t) dt$, so if I can bound this quantity with $0$, due to the fact that the norm is always positive or $0$, by the sandwich rule, I will be capable of saying that $||g_{n}||^{2}\to 0$, finishing the exercice. Nevertheless, I haven't been able to prove this... I have tried to use the fact that $\sin^{2n}(t)$ is in $[0,1]$, as $t\in[0,1]$... but I am quite stuck...  If anyone could give some hints or whatever I would be so grateful. Thanks in advanced!

Since $f$ is a continuous defined on a compact interval it attains its supremum $M$ and so we have $f(t) \le  M$, and for all $n \in \mathbb N$ it holds that $f\left(\frac{t}{n}\right) \le f(t) \le M$, since its graph  gets "streched" in the $x$-direction with the origin (or the $y$-axis) fixed. Moreover, since $\sin t < 1$ on $[0,1]$ then we have $\sin^n t \to 0$ as $n \to \infty$ and the conclusion follows
$$f\left(\frac{t}{n}\right)\sin^n t \le  M\sin^n t \to 0.$$
