Does $(f_n)=(n\sin(\frac{x}{n})-x)$ converge uniformly on $[-a,a]$ for $a\geq0$? I'm trying to solve the next problem: Let $\left(f_{n}\right)_{n\in\mathbb{N}}$
be a sequence of functions such that $f_{n}\colon\mathbb{R}\to\mathbb{R}$
is given by $f_{n}\left(x\right)=n\sin\left(\frac{x}{n}\right)-x$,
for all $n\in\mathbb{N}$.
$1.$ Prove that $\left(f_{n}\right)_{n\in\mathbb{N}}$ is pointwise
convergent.
$2.$ Prove that $\left(f_{n}\right)_{n\in\mathbb{N}}$ doesn't converge
uniformly.
$3.$ Does $\left(f_{n}\right)_{n\in\mathbb{N}}$ converge uniformly
on $\left[-a,a\right]$ for $a\geq0$?
For $1.$ I proved that $\left(f_{n}\right)_{n\in\mathbb{N}}$ coneverges
pointwise to 0. Given $x\in\mathbb{R}$, using L'Hopital's rule we
have that:
$$ \lim_{n\to\infty}\frac{\sin(\frac{x}{n})}{\frac{1}{n}}=\lim_{n\to\infty}\frac{\cos(\frac{x}{n})\cdot (-\frac{x}{n^{2}})}{-\frac{1}{n^{2}}}=x\cdot\lim_{n\to\infty}\cos(\frac{x}{n})=x\cdot\cos(0)=x.
$$
And therefore $\lim_{n\to\infty}f_{n}\left(x\right)=\lim_{n\to\infty}n\sin\left(\frac{x}{n}\right)-\lim_{n\to\infty}x=x-x=0.$
Hence $\left(f_{n}\right)_{n\in\mathbb{N}}$ converges pointwise to
0.
In $2.$ I took $\varepsilon=1$. Then given $n\in\mathbb{N}$, we
can take $x=n\pi\in\mathbb{R}$ and we have that 
$$
\mid f_{n}\left(x\right)-0\mid=\mid n\sin\left(\frac{n\pi}{n}\right)-n\pi\mid=n\pi>1=\varepsilon.
$$
And with this we can conclude that $\left(f_{n}\right)_{n\in\mathbb{N}}$
doesn't converge uniformly in $\mathbb{R}$.
I don't know for $3.$ if the answer is yes or no. I'm thinking that
may be the answer could be yes, since we are taking a compact set
and $f_{n}$ is continuous for all $n\in\mathbb{N}$, but I couldn't
prove this. Could you help me or give me some idea for this problem?
Thanks.

Note that, assuming that $x\in[-a,a]$,$$\left\lvert n\sin\left(\frac xn\right)-x\right\rvert=\left\lvert x\frac{\sin\left(\frac xn\right)}{\frac xn}-x\right\rvert=\lvert x\rvert\cdot\left\lvert\frac{\sin\left(\frac xn\right)}{\frac xn}-1\right\rvert\leqslant a\left\lvert\frac{\sin\left(\frac xn\right)}{\frac xn}-1\right\rvert.$$Now, use the fact that $\lim_{t\to0}\frac{\sin t}t=1$ and that$$(\forall x\in[-a,a]):\left\lvert\frac xn\right\vert\leqslant\frac an.$$
