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September 2nd, 2012, 02:41 AM   #1
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How to find lim as x -> inf. of sin(x) algebraically

Given:

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.
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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 ,

then for every sequence such that , it holds that .

But choosing and , so the limit does not exist.

[/color]
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September 2nd, 2012, 07:45 AM   #3
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Re: How to find lim as x -> inf. of sin(x) algebraically

Perhaps of interest,

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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,

The limit is 0 and can be it is possible to prove it using squeeze theorem.
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September 2nd, 2012, 08:33 AM   #5
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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,

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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!
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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,

Here's the proof:





By squeeze theorem, the result follows.
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