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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.

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  • $\begingroup$ Your question is a little unclear. Are we supposed to determine which of the four statements is correct? $\endgroup$ – Simon Hayward Nov 23 '12 at 12:54
  • $\begingroup$ Note that $f_n(0)=0$, which eliminates some of the potential answers immediately. $\endgroup$ – Mark Bennet Nov 23 '12 at 13:12
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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 $$\displaystyle\lim_{t \to 0}\frac{\sin t}{t}=1.$$

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  • $\begingroup$ From the above discussion,it appears that option $(c)$ is the correct choice. $\endgroup$ – learner Apr 3 '13 at 13:07
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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 $$ x = f'(0) = \lim_{n\to\infty}\frac{f(1/n) - f(0)}{1/n} = \lim_{n\to\infty} n \cdot \sin\frac xn $$

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  • $\begingroup$ This is the first time that I see this notation $f\colon y \mapsto \sin(yx)$. Did not you mean $f(y)=\sin(yx)$? $\endgroup$ – allesia_b Aug 25 '18 at 1:29
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You can use l'Hopital's rule to evaluate the limit

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

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  • $\begingroup$ Thanks........... $\endgroup$ – Cloud JR K Nov 28 '18 at 20:16

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