Find the set S on which $f_n(x)=n\sin \frac{x}{n}$ converges pointwise, and find the limit function. The sequence converges to $x$ since the Maclaurin expansion of $\sin x$ is $\sum_{k=0}^\infty(-1)^{k} \frac{x^{2k+1}}{(2k+1)!}$ , hence
$$n\sin \frac xn = \sum_{k=0}^\infty \frac{(-1)^kx^{2k+1}}{(2k+1)!n^{2k}} = x + \sum_{k=1}^\infty \frac{(-1)^kx^{2k+1}}{(2k+1)!n^{2k}}.$$
Thus $n\sin\frac{x}{n}\rightarrow x$ as $n\rightarrow \infty$. Now, the manual to Trench's book "Introduction to Real Analysis" does a slightly different approach that I cannot grasp yet:
"From Taylor's theorem, $\sin\frac{x}{n}=\frac{x}{n}-cos\theta(x,n)\frac{x^3}{6n^3}$, where $\cos\theta(x,n)$ is between zero and $\frac{x}{n}$. Therefore $\left |f_n(x)-x \right |\leq \frac{x^2}{6n^2}$, so $\lim_{n \to \infty}f_n(x)=x, -\infty< x< \infty$".
Where did the cosine of theta come from? Is it a trick to make the absolute value less than or equal, rather than just equal?
Best Answer
I think this is all rather roundabout. We know that $\sin'=\cos$, so for fixed $x$ and $f_x:y\mapsto\sin(xy)$ one has $\lim_{y\to0}\frac{\sin(xy)}y=f_x'(0)=\cos(0)x=x$, and in particular $\lim_{n\to\infty}\frac{\sin(x/n)}{1/n}=x$.