Behaviour of $f_n(x) = \sin (\frac{x}{n})$ for each $x ∈ \Bbb{R} $ and $n ∈ \Bbb{N}$

Question

Consider $$f_n(x)=\sin\Big(\frac{x}{n}\Big)$$ on $\Bbb{R}$.

• Calculate $f(x) = \lim_{n→∞} f_n(x)$ and determine the domain of $f(x)$

• Is $f_n(x)$ uniformly converging to $f(x)$ on $[−π, π]$ ?

• Is $f_n(x)$ uniformly converging to $f(x)$ on $\Bbb{R}$ ?

Answer 1

Point-wise limit is obviously $$f(x) = 0$$ On $[-\pi, \pi]$ function is uniformly converging because $$f_n(x) - f(x) \leq sin(\frac{\pi}{n})$$ so the distance beetwen functions is as small as you wish. It'is not uniformly converging on $\mathbb{R}$, if would, lets choose $\varepsilon = \frac{1}{2}$. By definition there is some $N$ that for all $n > N$, $f_n(x) - f(x) < \varepsilon$ for all $x$. But clearly just take $n = 3N$ and you'll get $$f_{3N}(\pi) - f(\pi) = sin(\frac{3N\pi}{N}) = sin(3\pi) = 1 > \frac{1}{2}$$ Contradiction.