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 September 2nd, 2012, 02:41 AM #1 Member   Joined: Aug 2012 Posts: 88 Thanks: 0 How to find lim as x -> inf. of sin(x) algebraically Given: $\lim_{x \to \infty } \sin x$ I understand that sin(x) oscillates between -1 and 1 as we go towards infinity in either direction, but if I didn't know what the graph looked like or the general behavior of sin(x), how would I write algebraically to show that the limit doesn't exist? The work I show is usually plugging in large numbers close to each other to show that the y-values of sin(x) oscillate, but I don't think that is a valid way of doing this unless I use like 10 different x-values and plug them all in.
September 2nd, 2012, 03:27 AM   #2
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Re: How to find lim as x -> inf. of sin(x) algebraically

[color=#000000]Suppose that $\lim_{x\to\infty}\sin(x)=\ell\in\mathbb{R}$,

then for every sequence $\{x_{n}\}\in\mathbb{R}$ such that $x_{n}\to\infty$, it holds that $\lim_{n\to\infty}\sin(x_n)=\ell$.

But choosing $x_n=2\pi n \Rightarrow \sin(2\pi n )\to 0$ and $x_{n}=2\pi n +\frac{\pi}{2}\Rightarrow x_{n}\to 1$, so the limit does not exist.

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 September 2nd, 2012, 07:45 AM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,935 Thanks: 1129 Math Focus: Elementary mathematics and beyond Re: How to find lim as x -> inf. of sin(x) algebraically Perhaps of interest, $\lim_{x\to\infty}\,\frac{\sin(x)}{x}$
September 2nd, 2012, 07:51 AM   #4
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Re: How to find lim as x -> inf. of sin(x) algebraically

Quote:
 Originally Posted by greg1313 Perhaps of interest, $\lim_{x\to\infty}\,\frac{\sin(x)}{x}$
The limit is 0 and can be it is possible to prove it using squeeze theorem.

 September 2nd, 2012, 08:33 AM #5 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,935 Thanks: 1129 Math Focus: Elementary mathematics and beyond Re: How to find lim as x -> inf. of sin(x) algebraically Claim: Let f(x) be a continuous function bounded from above and below by a and b, with both a and b finite, for all x. Then, $\lim_{x\to\infty}\,\frac{f(x)}{x}\,=\,0$
September 2nd, 2012, 08:36 AM   #6
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Re: How to find lim as x -> inf. of sin(x) algebraically

Quote:
 Originally Posted by PhizKid . . . I don't think that is a valid way of doing this unless I use like 10 different x-values and plug them all in.
That is still not valid!

September 2nd, 2012, 08:38 AM   #7
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Re: How to find lim as x -> inf. of sin(x) algebraically

Quote:
 Originally Posted by greg1313 Claim: Let f(x) be a continuous function bounded from above and below by a and b, with both a and b finite, for all x. Then, $\lim_{x\to\infty}\,\frac{f(x)}{x}\,=\,0$
Here's the proof:

$a \, < \, f(x) \, < \, b$

$\Rightarrow \frac{a}{x} \, < \, \frac{f(x)}{x} \, < \, \frac{b}{x}$

By squeeze theorem, the result follows.

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