Uniform convergence of $\{n\sin\frac{1}{nx}\}$ on $(0,1]$ Show that $f_n(x)=n\sin\frac{1}{nx}$ is not converges uniformly to limit function on $(0,1]$.
Tell me if that is right:
I have found that the limit function of $f_n$ is $\frac{1}{x}$.
Now, 
$$|n\sin\frac{1}{nx}-\frac{1}{x}|\leq |n\sin\frac{1}{nx}|+\frac{1}{x}$$
Since, $\sin x< x$ for small $x\in (0,1]$ we have that:
$$|n\sin\frac{1}{nx}-\frac{1}{x}|+\frac{1}{x}<n\frac{1}{nx}+\frac{1}{x}=\frac{2}{x}$$
Thus, exist $\epsilon>0$, (i.e $\epsilon=2$) s.t for every $N\in\mathbb{N}$ and $n>N$:
$$|n\sin\frac{1}{nx}-\frac{1}{x}|>2$$
Hence $f_n \to f$ not uniformly. 
 A: The non-uniform convergence of a sequence is normally proved in the following way.
Uniform convergence is

For any fixed $\epsilon>0$ there exists an $N$, such that for all $n>N$
  $$
|f_n(x) - f(x)| < \epsilon \quad \text{for any } x,$$
  or  more simply put 
  $$
\sup_{x} |f_n(x) - f(x)|  \to 0.
$$

Now if we can negate above argument and try to find something like

For any fixed $N$, such that there exists an $\epsilon>0$ and $n > N$:
  $$
|f_n(x) - f(x)| \geq \epsilon \quad \text{for some }x,\tag{1}
$$
  or more simply put 
  $$\sup_{x} |f_n(x) - f(x)| \not \to 0.\tag{2}$$

Then the convergence is not uniform.
For (1), there are many choices of such $x$ you can try (there are many other choices, based on an inverse involving an integer times $\pi$): for any $N$ fixed, let
$$
x = \frac{1}{N\pi},
$$
then letting $n= 2N > N$
$$
\left|n\sin \frac{N\pi}{n} - N\pi\right| = N(\pi-2) \geq \epsilon.
$$
Done.
Or you could try (2)
$$
\sup_{x}  \left| n\sin \frac{1}{nx}- \frac{1}{x}\right| \geq  \left| n\sin \frac{1}{n(2/(n\pi))}- \frac{1}{2/(n\pi)}\right| = n(\pi/2-1)\not\to 0 \quad \text{ as } n\to \infty.
$$
