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My Question is:

  • Examine the continuity of $$f(x) = \lim_{n \to \infty} \frac{x}{(2\sin{x})^{2n}+1} \qquad \text{for} \ x \in \mathbb{R}$$

How can I do this? Honestly, speaking I have $\text{no idea}$ of doing, this as this seems to be a tough limit. What I only know is that as $n \to \infty$ $\sin{n}$ $\text{oscillates}$ between $-1$ and $1$.

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up vote 14 down vote accepted

The key here lies in the longterm behavior of $(2\sin x)^{2n}$. Note that $\sin x$ is fixed: the limit is over $n$, so what is changing is the exponent. So we are really looking at the behavior of $a^{2n}$ as $n\to\infty$, where $a$ is some number between $-2$ and $2$ (namely, $a=2\sin x$).

If $-1\lt 2\sin x\lt 1$, then $(\sin x)^{2n}\to 0$ as $n\to\infty$, so $f(x) = x$. On the other hand, if $2\sin x\gt 1$ or $2\sin x\lt -1$, then $(2\sin x)^{2n} = \left((2\sin x)^2\right)^n$ will go to $\infty$ as $n\to\infty$, so the limit will be $0$. Finally, if $2\sin x = \pm 1$, then $(2\sin x)^{2n}\to 1$ as $n\to\infty$, so the limit will be $1$. That is: $$\lim_{n\to\infty}\frac{x}{(2\sin x)^{2n}+1} = \left\{\begin{array}{lll} x &&\text{if }|\sin x|\lt \frac{1}{2}\\ \frac{x}{2} &\quad&\text{if }|\sin x|=\frac{1}{2};\\ 0 && \text{if }|\sin x|\gt\frac{1}{2}. \end{array}\right.$$ Is this enough for you to finish off the problem, analyzing the continuity of $f(x)$? If not, I'll add more.

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Proceeding from where Arturo left note that:

  • $|2\sin{x}|< 1$ $\Longrightarrow$ $x \in \displaystyle\Bigl(n\pi-\frac{\pi}{6},n\pi+\frac{\pi}{6}\:\Bigr)$

  • $|2\sin{x}|=1 \Longrightarrow \displaystyle x =n\pi+\frac{\pi}{6}$

  • $|2\sin{x}|>1 \Longrightarrow \displaystyle x \in \Bigl(n\pi + \frac{\pi}{6}, n\pi+\frac{5\pi}{6}\:\Bigr)$

So your $f(x)$ is basically, $$ f(x) = \left\{\begin{array}{lll} x & \text{if}\ \Bigl(n\pi-\frac{\pi}{6},n\pi+\frac{\pi}{6}\Bigr) \\\ \frac{x}{2} & \text{if} \ x =n\pi+\frac{\pi}{6} \\\ 0 & \text{if} \ x \in \Bigl(n\pi + \frac{\pi}{6}, n\pi+\frac{5\pi}{6}\Bigr)\end{array}\right.$$

It is easy to note $\text{that the only points which gives us trouble are}$ $x =n\pi \pm \frac{\pi}{6}$. Now your job is to check the continuity of $f(x)$ at $x=n\pi \pm \frac{\pi}{6}$.

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@Chandrasekhar: Spoon feeding is not helpful for anybody! What Arturo had was enough. – JavaMan Aug 13 '11 at 3:41
@DJC: Discussions about spoon-feeding aren't helpful, either! :) – The Chaz 2.0 Aug 13 '11 at 7:04
@Chaz, Chandrasekhar: Discussions about spoon-feeding and general homework policies are indeed worthwhile though they should not occur here. I will gladly bring up this issue in a discussion next time. – JavaMan Aug 13 '11 at 16:29
@DJC: Such discussions are probably better held in meta, though I suspect not much will come out of them. For the record, I was somewhat annoyed/dismayed that Chandru would not give the OP some time to try to work it out for himself before spilling most of the beans. – Arturo Magidin Aug 13 '11 at 18:13
@ABD: The truly ideal way, I think, would have been for you to work a solution based on where you were, and post either an answer to your own question or perhaps better an addendum to your original question containing your attempt at solution. Then everyone could have commented and that solution to point out any possible problems, or to let you know it was okay. If you were having trouble figuring out the values of $x$ for which $\sin x$ had the required values... well, that's why I said if what I wrote was not enough I could add. In short, you can always ask follow-ups. – Arturo Magidin Aug 13 '11 at 18:15

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