Does $u_n(x) = n \sin(x/n)$ converges uniformly on $[-5,7]$? I have to show that my $u_n(x)= n \sin\frac{x}{n}$ converges uniformly on $[-5,7]$.
My u(x) = $\lim\limits_{n \to\infty} n \sin\frac{x}{n} = x$, so I'm trying to find my sup |$ n \sin\frac{x}{n} - x$| on $[-5,7]$. 
What I wanted to use is differential calculus $\frac{\text{d}}{\text{d}x}(n \sin\frac{x}{n}-x) = \cos\frac{x}{n}-1$  $\cos\frac{x}{n} - 1 \to x= 2kn\pi, k \in \mathbb{Z}. $ 
$$u_n(2n\pi) = n \sin\frac{2n\pi}{n}-2n\pi= $$
$$=n \sin2\pi - 2n\pi$$ 
And the endpoints: 
$$u_n(-5) = n \sin\frac{-5}{n} +5 $$ 
$$u_n(7) = n \sin\frac{7}{n} -7 $$
And I'm stuck here. I don't know how to determine whether it's uniformly convergent or not. Can somebody help me? 
 A: Take $\varepsilon>0$. Since $\lim_{x\to0}\frac{\sin x}x=1$, there is some $\delta>0$ such that$$|x|<\delta\implies\left|\frac{\sin x}x-1\right|<\frac\varepsilon7.$$Now, take $N\in\Bbb N$ such that $\frac1N<\frac\delta7$. Then, for each $x\in[-5,7]$ and each $n\geqslant N$, $\left|\frac xn\right|<\delta$ and therefore$$\left|\frac{\sin\left(\frac xn\right)}{\frac xn}-1\right|<\frac\varepsilon7.$$So,$$\left|n\sin\left(\frac xn\right)-x\right|<\frac\varepsilon7|x|\leqslant\varepsilon.$$
A: We begin with the inequality
$$ |\sin(t)-t|\le \frac16|t|^3$$
Letting $t=x/n$, we find that  
$$\begin{align}
|n\sin(x/n)-x|\le \frac{|x|^3}{6n^2}
\end{align}$$
If $|x|\le 7$, then for any $\varepsilon>0$,
$$|n\sin(x/n)-x|<\varepsilon$$
whenever $n>N=\sqrt{\frac{7^3}{6\varepsilon}}$.


NOTE:

The OP found that $\sup_{x\in [-5,7]}|n\sin(x/n)-x|=|n\sin(7/n)-7|$.  Then, applying the inequality $|\sin(t)-t|\le \frac16 t^3$, we find again that for any given $\varepsilon>0$
$$\sup_{x\in [-5,7]}|n\sin(x/n)-x|\le \frac{7^3}{6n^2}<\varepsilon$$
whenever $n>n=\sqrt{\frac{7^3}{6\varepsilon}}$.
