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anonymous

  • 5 years ago

Find the limit: x to infinity xsin(2/x)

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  1. anonymous
    • 5 years ago
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    Break it up in parts lim x and lim sin 2/x. For sin 2/x, look in your notes for what happens when a number say a is over in finity. Therefore it becomes sin of that quantity.

  2. anonymous
    • 5 years ago
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    however i am not getting the answer by applying the limit the following way: |sin(2/x)|<= 1 so |xsin(2/x)|<=|x||sin2/x| triangle inequality now if i apply limit, lim |x|=infinity since it diverges, the smaller ones are bound to diverge. but in effect the answer is 2

  3. anonymous
    • 5 years ago
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    How do you get 2, rsaad2?

  4. anonymous
    • 5 years ago
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    can someone explain as to why? you use taylor polynomial, you get 2, but with comparison test you get infinity. confused ;S

  5. anonymous
    • 5 years ago
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    Why are you going into higher mathematics? Why not Cal I limits?

  6. anonymous
    • 5 years ago
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    as in?

  7. anonymous
    • 5 years ago
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    like i said i am using simple limit, i am getting infinity.

  8. anonymous
    • 5 years ago
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    Saw a similar example in my text book. Looking it up, I'll get back...

  9. anonymous
    • 5 years ago
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    okay. here it is lim x (2/x) {sin(2/x)/ (2/x)} lim 2 lim {sin(2/x)/ (2/x)} 2*1 2

  10. anonymous
    • 5 years ago
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    and the reason as to why sin(2/x)/(2/x) = 1 is because its similar to sin x/x when x-> 0 so lim sin x/x = 1.

  11. anonymous
    • 5 years ago
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    Yeah, I was researching for my own learning. The proof is let u =sin 2x, then x = 2/u; limits changed to u goes to zero. It becomes limit as u goes to zero of (2 sin u)/u. Of course (sin u)/u is an identity equal to one. Therefore answer two. Good job rsaad2.

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