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0 times sin(infinity) Watch

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    If you are asked to evaluate the limit as x tends towards 0 of x[sin(1/x)], you would get 0 times sin(infinity) with quite basic algebraic manipulation. Now of course, we know that sin(infinity) does not exist. My question is, would it be mathematically incoherent to neglect that fact, and nonetheless proceed to say that 0 times anything is zero, regardless of the fact that said anything does not exist? I know that the answer of zero is correct through the use of the squeeze theorem, but my question is about the nature of multiplication through zero.
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    This is probably in the wrong section, I would appreciate it if someone could move it to that appropriate.
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    (Original post by Hubrillity)
    If you are asked to evaluate the limit as x tends towards 0 of x[sin(1/x)], you would get 0 times sin(infinity) with quite basic algebraic manipulation. Now of course, we know that sin(infinity) does not exist. My question is, would it be mathematically incoherent to neglect that fact, and nonetheless proceed to say that 0 times anything is zero, regardless of the fact that said anything does not exist? I know that the answer of zero is correct through the use of the squeeze theorem, but my question is about the nature of multiplication through zero.
    It is indeed incoherent. Let's try applying your argument to a different example. Consider the limit as x tends towards 0 of x(1/x). Subbing in 0, we get 0 times (1/0) = 0 times something = 0. But obviously we know that x(1/x) is constant (when defined) and the limit is 1.
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    (Original post by Hubrillity)
    If you are asked to evaluate the limit as x tends towards 0 of x[sin(1/x)], you would get 0 times sin(infinity) with quite basic algebraic manipulation. Now of course, we know that sin(infinity) does not exist. My question is, would it be mathematically incoherent to neglect that fact, and nonetheless proceed to say that 0 times anything is zero, regardless of the fact that said anything does not exist? I know that the answer of zero is correct through the use of the squeeze theorem, but my question is about the nature of multiplication through zero.
    0\times \sin (\infty) is a meaningless piece of notation.

    \lim_{x\to 0} x\sin (1/x) has a very specific meaning and is equal to 0
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    (Original post by 13 1 20 8 42)
    It is indeed incoherent. Let's try applying your argument to a different example. Consider the limit as x tends towards 0 of x(1/x). Subbing in 0, we get 0 times (1/0) = 0 times something = 0. But obviously we know that x(1/x) is constant (when defined) and the limit is 1.
    That's perfect, thank you very much.
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    Wait, also:

    Would a different way to look at it be this?

    As x approaches zero, the limit of sin(1/x) remains undefined while the function gets faster, whereas the limit of y=x itself is zero. So instead of solving this algebraically, we can solve it graphically. The function of sin (1/x) is squeezed into a tendency towards 0 by the the function of y=x, thus forcing the limit of x sin(1/x) to be 0. Am I going wrong anywhere?
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    (Original post by Zacken)
    0\times \sin (\infty) is a meaningless piece of notation.

    \lim_{x\to 0} x\sin (1/x) has a very specific meaning and is equal to 0
    ^This

    (Original post by 13 1 20 8 42)
    It is indeed incoherent. Let's try applying your argument to a different example. Consider the limit as x tends towards 0 of x(1/x). Subbing in 0, we get 0 times (1/0) = 0 times something = 0. But obviously we know that x(1/x) is constant (when defined) and the limit is 1.
    1/x is not bounded, but sin(1/x) is.
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    (Original post by RogerOxon)
    ^This


    1/x is not bounded, but sin(1/x) is.
    True. But he didn't mention the boundedness in his argument, so I did not see it as necessary to use a bounded function to show the fault with said argument. However I would concede that Zacken's point is more important, people often misuse infinity like this.
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    (Original post by Hubrillity)
    Wait, also:

    Would a different way to look at it be this?

    As x approaches zero, the limit of sin(1/x) remains undefined while the function gets faster, whereas the limit of y=x itself is zero. So instead of solving this algebraically, we can solve it graphically. The function of sin (1/x) is squeezed into a tendency towards 0 by the the function of y=x, thus forcing the limit of x sin(1/x) to be 0. Am I going wrong anywhere?
    It seems you're essentially just giving a geometric interpretation of the squeeze/sandwich theorem. True enough, but not really formal enough to serve as a proof.
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    Thanks guys, this helps lots. I might have a few more questions in regards to limits so I'll keep posting them here if anyone has free time to answer them.
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    (Original post by Hubrillity)
    Wait, also:

    Would a different way to look at it be this?

    As x approaches zero, the limit of sin(1/x) remains undefined while the function gets faster, whereas the limit of y=x itself is zero.
    sin(1/x) is defined for all values of x except x=0. I'm guessing what you mean is that it doesn't tend towards a defined limit (which is true).

    But that isn't actually relevant here. The key point is that |sin(1/x)| <=1 for all x\neq 0. This is what allows us to say that it gets 'squeezed to 0' by multiplying by x as x \to 0.
 
 
 
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