The sequence of functions $$f_n(x)=n \sin\left(\frac{x}{n}\right)$$ is pointwise convergent everywhere in the interval $(-\infty,\infty)$ and converges to the function $f(x)=x$.
This function $f(x)$ is continuous in the given interval but that does not tell us anything about the uniform convergence of the given sequence function in the given interval.
I try to find out the supremum of $|f_n(x)-f(x)|$ i.e.$|n \sin\left(\frac{x}{n}\right)-x|$ as $n$ tends to $\infty$.
Using derivative test for Maxima/minima, I get $\cos\left(\frac{x}{n}\right)=1$. This gives me $\frac{x}{n}=2p\pi$ where $p$ is an integer. Now, $x=2p\pi n$.
We get supremum as $\left|n \sin\left(\frac{2p\pi n}{n}\right)-2p \pi n\right|= 2p\pi n$.
This goes to $\infty$ as $n$ goes to $\infty$. Is this enough to prove that given function sequence is not uniformly convergent in given interval by application of $M_n$ test?