# 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.

• Your question is a little unclear. Are we supposed to determine which of the four statements is correct? Commented Nov 23, 2012 at 12:54
• Note that $f_n(0)=0$, which eliminates some of the potential answers immediately. Commented Nov 23, 2012 at 13:12

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 $$\displaystyle\lim_{t \to 0}\frac{\sin t}{t}=1.$$

• From the above discussion,it appears that option $(c)$ is the correct choice. Commented Apr 3, 2013 at 13:07

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 $$x = f'(0) = \lim_{n\to\infty}\frac{f(1/n) - f(0)}{1/n} = \lim_{n\to\infty} n \cdot \sin\frac xn$$

• This is the first time that I see this notation $f\colon y \mapsto \sin(yx)$. Did not you mean $f(y)=\sin(yx)$?
– user486983
Commented Aug 25, 2018 at 1:29

You can use l'Hopital's rule to evaluate the limit

$$\lim_{n \to \infty} n\sin n/x = \lim \frac{\sin x/n}{1/n} = \lim \frac{\cos(x/n) (-x/n^2)}{-1/n^2} = \lim x \cos x/n = x$$

• Thanks........... Commented Nov 28, 2018 at 20:16