Is $f_n(x)=\frac{x}{n}\sin\frac{x}{n}$ uniform convegent

Question

Consider the functions

$f_n:\mathbb{R}\rightarrow\mathbb{R}$

for $n=1,2,...$ defined by $f_n(x)=\frac{x}{n}\sin\frac{x}{n}$

Give all the points in $\mathbb{R}$ where $f_n$ converges poinwisely

does ${f_n}$ converge uniformly on $\mathbb{R}$

I think since $f_n(0)=0$

then $\lim f_n=0$ as

$n \rightarrow \infty$

But I am not sure how you would show it is uniform convergent.

Answer 1

The function convege pointwise to $0$ for all $x\in \mathbb{R}$ because $$\forall x\in \mathbb{R} |f_{n}(x)|=\Bigg| \frac{x}{n} \cdot \sin \Big( \frac{x}{n} \Big) \Bigg| \leq \Big| \frac{x}{n} \Big| \rightarrow_{n\to \infty}0 $$ However there no uniform convegent becaue for e.g $\epsilon=1$ for all $n\in \mathbb{N}$ we can choose $x_0 := \frac{n\pi}{2}$ : $$ f_{n}(x)= \frac{ \frac{n\pi}{2} }{n} \cdot \sin \Big( \frac{ \frac{n\pi}{2} }{n} \Big) = \frac{\pi}{2} >1 $$ So $f_n\rightarrow 0$ but not uniformly.