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I'm trying to prove that the limit of sin x as x->infinity is not equal to 1/2. I know that this is true, but I can't seen to figure out how to prove it using the precise definition of a limit.

What I have so far is this,

e>0, M>0

abs(sin x - 1/2)<e whenever x>M

I also think that I can use the fact that,

abs(sin x)<=1 for all x

But I don't know for sure. I've also seen some really extensive proofs for stuff like sin x/x (ref 1). So I may be doing this completely wrong.

Thanks for any help.

  1. http://tutorial.math.lamar.edu/Classes/CalcI/ProofTrigDeriv.aspx
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3 Answers 3

up vote 1 down vote accepted

(Do see comments bellow.)

Well, for this to be true, it must hold that

$$\forall \epsilon > 0 \exists \delta > 0 \forall x \in P^-(\infty, \delta): |\sin x| < \epsilon.$$

Obviously, this does not hold for $x \in \{\frac\pi{2} + 2k\pi, k\in \mathbb{Z}\}$, for example. $\sin x$ will always reach the value of $1$ in those points.

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I think I understand what you mean by this. But my suggestion would be to explain it in as simple a manner as possible. Because I'm not familiar with some of the symbols. –  ASAAR Sep 9 '13 at 10:00
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Well, if the limit of $\sin x$ in infinity is supposed to equal to zero, it means that I can choose some arbitrarily small $\epsilon$ such that for all points around infinity, the value of $\sin x$ will be no more than epsilon. "All points around infinity" are defined as $x \in (\frac1\epsilon, \infty)$. There are infinitely many points in this interval; some of which are equal to $\frac\pi{2} + 2k\pi$ for some $k$. But at these points, the value of $\sin x$ is 1. Therefore, the limit cannot be zero. –  David Čepelík Sep 9 '13 at 10:06
    
I understand how this would apply now. But your particular proof isn't for "sin x as x->infinity is not equal to 1/2", is it? It seems more fitting for "sin x as x->infinity is not equal to 0". –  ASAAR Sep 9 '13 at 10:14
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Oh, I am sorry, I missed that $\frac12$. It will be the same: for $\epsilon < \frac12$, for all points around infinity it has to hold that $|\sin x - \frac12| < \epsilon$. But it does not. –  David Čepelík Sep 9 '13 at 10:22
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$P$ is the neighborhood of a given point. When finding the limit in $c \in \mathbb{R}^*$, we claim that some statement must hold for all points that are close to $c$, or better, at most $\delta$ apart from $c$. The set of "all points around $c$ no farther than $\delta$" is then $P(c, \delta)$ with the minus sign denoting we only examine the points to the left of $c$ (as it makes no sense to consider points to the right of $c = \infty$ in this particular case). Note that $c$ itself does not belong to $P(c, \delta)$. –  David Čepelík Sep 9 '13 at 10:43

Choose $\epsilon_0 = 1/4$. Then for any $M>0$, choose an integer $K$ such that $K\pi>M$. Then $$|\sin(K\pi)-1/2|=1/2>1/4=\epsilon_0.$$ Thus by the definition of limit, $$\displaystyle\lim_{x\to\infty}\sin x\neq 1/2.$$

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Consider the following theorem: If $\lim_{x\to \infty } f(x)=L$ and $x_n$ is a sequence of numbers such that $\lim _{n\to \infty }x_n =\infty $, then $\lim_{n\to \infty }f(x_n)=L$. If you feel comfortable using this theorem, then you can now prove that $\lim_{x\to \infty }\sin(x)\ne \frac{1}{2}$ by considering a suitable sequence $x_n$ with $\lim_{n\to \infty }x_n=\infty $ and $\sin(x_n)=1$ for all $n$. It then follows that $\lim_{n\to \infty }\sin (x_n)=1\ne \frac{1}{2}$, and thus proves the claim.

If you haven't seen this theorem before, try proving it. If you want to give a proof that is as close as possible to directly using the definition of limit, then try to figure out why the above works. Choose $\epsilon = \frac{1}{4}$ for instance, and show that for all $M$ there exists $x>M$ with distance between $\sin(x)$ and $\frac{1}{2}$ bigger than $\epsilon $.

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I like where you went with this. But I'm definitely going to need to try and prove that theorem to myself tonight before I run with that method. –  ASAAR Sep 9 '13 at 10:17

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