The function $f_n(x) = n \sin(x/n)$ The function $f_n(x) = n \sin(x/n)$. Then which option is the correct?
(a) does not converge for any $x$ as $n \to\infty$.
(b) converges to the constant function $1$ as $n \to\infty$.
(c) converges to the function $x$ as $n \to\infty$.
(d) does not converge for all $x$ as $n \to\infty$.
If $x=n\pi$ then function will be zero. But what should be the general case? Thanks for help.
 A: You want to find $\displaystyle\lim_{n \to \infty}f_n(x)$ for $x$ in some $E$.
Take a fixed $x \in E$ (you cannot take $x=n\pi$) and evaluate $\displaystyle\lim_{n \to \infty}f_n(x)$.
Use that $$\displaystyle\lim_{t \to 0}\frac{\sin t}{t}=1.$$
A: You can use l'Hopital's rule to evaluate the limit
$$
\lim_{n \to \infty} n\sin n/x = \lim \frac{\sin x/n}{1/n} = \lim \frac{\cos(x/n) (-x/n^2)}{-1/n^2} = \lim x \cos x/n = x
$$
A: Note that
$$ \sin x = \sum_{k=0}^\infty(-1)^{k} \frac{x^{2k+1}}{(2k+1)!} $$
Hence 
$$ n\sin \frac xn = \sum_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!n^{2k}}  = x + \sum_{k=1}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!n^{2k}} $$
No the last sum converges termwise and hence (by dominated convergence) also as a sum) to $0$ for $n \to \infty$, we have 
$$ n \sin \frac xn \to x, \qquad n\to \infty $$

Another way to see this is to consider $f\colon y \mapsto \sin(yx)$, then $f'(y) = x\cos(yx)$ and 
$$ x = f'(0) = \lim_{n\to\infty}\frac{f(1/n) - f(0)}{1/n} = \lim_{n\to\infty} n \cdot \sin\frac xn $$
