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Simple Limit

  1. Jul 30, 2008 #1
    Asked to compute:

    \lim_{n\to\infty} (-1)^nsin(1/n)

    I've broken this limit down into:

    \lim_{n\to\infty} (-1)^n * \lim_{n\to\infty}sin(1/n)

    I've determined [tex] \lim_{n\to\infty}sin(1/n) = 0[/tex]

    Now I have [tex] \lim_{n\to\infty} (-1)^n * 0[/tex]

    This is where I run into trouble...

    Attempting to solve for [tex] \lim_{n\to\infty} (-1)^n [/tex]:

    -I've tried plugging in integers and rational numbers for n. It jumps to -1 and 1 with integers, and spits out complex numbers when I plug in rational numbers.
    -I've also tried graphing this function on a calculator to no avail.
    -I've also plugged it into maple and it spits out: (-1..1).

    Is it safe to say [tex] \lim_{n\to\infty} (-1)^n [/tex] does not exist?

    In which case, I have something that does not exist multiplied by 0, and anything multiplied by 0 equals 0... but I have "nothing" not "anything" ;)
  2. jcsd
  3. Jul 30, 2008 #2


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    Staff Emeritus
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    Yes, [itex]\lim_{n\rightarrow \infty} (-1)^n[/itex] does not exist. However,
    [tex]- sin(1/n)\le (-1)^nsin(1/n)\le sin(1/n)[/itex]
    and sin(1/n) goes to 0 as n goes to infinity. What does that tell you?
  4. Jul 30, 2008 #3
    Oh I see...sammich' theorem!
    How do you produce that inequality?
    I'm not sure how to work with sin(1/x)... can it be related to sin(x) somehow?
  5. Jul 30, 2008 #4
    For [tex]h(x) \leq f(x) \leq g(x) [/tex]

    where [tex]\lim_{n\to\infty} h(x) = L,\lim_{n\to\infty} g(x)=L[/tex]


    [tex]\lim_{n\to\infty} f(x)=L[/tex]

    Look at HallofIvy's hint. What is L?
  6. Jul 30, 2008 #5
    Yes I'm aware that the limit is sandwiched between two 0s and is therefore 0. I guess what I am asking for is a proof of the inequality. How did you decide to pick sin(1/n)? How do I know that it is between those functions?
  7. Jul 30, 2008 #6


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    What? Your question was about (-1)n sin(1/n). n is either even or odd. (-1)n is either 1 or -1. I "picked" sin(1/n) because that was the function multiplied by (-1)n.
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