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anonymous

  • 5 years ago

What is the limit of (x Sin (1/x)) as x becomes positive infinite?

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  1. anonymous
    • 5 years ago
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    i believe the limit is 0 because as x increases, 1/x decreases to where it is approaching zero and the sin (0) is 0. and anything multiplied by 0 is 0.

  2. anonymous
    • 5 years ago
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    Use a substitution u=1/x, then the limit will be \[\lim_{u \rightarrow 0}{\sin x \over x}=1\]

  3. anonymous
    • 5 years ago
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    it is 0

  4. anonymous
    • 5 years ago
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    Sorry \[\lim_{u \rightarrow 0} {\sin u \over u}=1\]

  5. anonymous
    • 5 years ago
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    limits is zero , its seen by pinching theorm

  6. anonymous
    • 5 years ago
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    Anwar..i think you might have to use L'Hopital's Rule.

  7. anonymous
    • 5 years ago
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    I don't think L'hopital's rule can work here. I still believe it's 1 by the substitution I used.

  8. anonymous
    • 5 years ago
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    check it, sub in a big number'

  9. anonymous
    • 5 years ago
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    it goes to zero

  10. anonymous
    • 5 years ago
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    the larger you're number gets, the closer you are to zero because it is a fraction.

  11. anonymous
    • 5 years ago
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    @thamir: When substituting \(u=1/x\), \(\sin (1/x)= \sin u\) and \(x=1/u\). Also the approaching point will be zero instead of infinity since \(1/\infty \rightarrow 0\).

  12. anonymous
    • 5 years ago
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    -1<sin(1/x) < 1 , this is known , multiply through by x ( it is positive because we are considering x->+infinity

  13. anonymous
    • 5 years ago
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    -x<xsin(1/x)<x

  14. anonymous
    • 5 years ago
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    wait, thats not going to work :|

  15. anonymous
    • 5 years ago
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    something similar lol

  16. anonymous
    • 5 years ago
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    @elecengineer: When you substitute a large number the x goes to \(\infty \) and the sin(1/x) goes to \(0\). And \( \infty \times 0 \ne0\).

  17. anonymous
    • 5 years ago
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    L'Hopitals Rule does work AnwarA and you are right the answer is 1. lim sin(1/x)/(1/x) =lim [(-1/x^2)cos(1/x)]/(-1/x^2) =lim cos(1/x) =cos(0)=1

  18. myininaya
    • 5 years ago
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    \[\lim_{x \rightarrow \inf}\frac{\sin(\frac{1}{x})}{\frac{1}{x}}\] let u=1/x x->inf, u->0 so the limit is 1

  19. anonymous
    • 5 years ago
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    Thanks AnwarA for solving it. Thanks chris777 for teaching me the L'Hopitals Rule. And last but not least thanks to elecengineer for trying to solve using the squeeze theorem. Now, I need to get the correct answer which is 1 by using the squeeze theorem. When trying I reached the same point where elecengineer reached. How can we solve it using the squeeze theorem?

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