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Limits for Sin(1/x^n)

  1. Oct 23, 2008 #1
    Hi

    I have been trying to work with limits and i came across a problem where it was defined that the limit for Sin(1/x) is not defined at x=0. Well i completely understand that part because the function oscillates for any small limit near 0. But i don't really understand how to calculate the limit for x for example at 0.1. Because the function still oscillates and what would be the ideal limit value that i need to take so that the function will proceed towards value sin(1/0.1)? And how to calculate similarly for sin(1/x^2), sin(1/x^3) etc until Sin(1/x^n) for x=0.1 from the definition of limits.

    I tried to do a matlab code and generate graph to see if there is a pattern in oscillation, but unfortunately i could not decode anything. I have attached the graph here, the blue curve indicates sin(1/x), red for sin(1/x^2), green for sin(1/x^3) and black for sin(1/x^4).

    P.S: In my opinion i felt this problem does not fall into homework questions, but i'm extremely sorry if someone else feels that way
     

    Attached Files:

  2. jcsd
  3. Oct 23, 2008 #2

    HallsofIvy

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    Sin(x) is a continuous function so
    [tex]\lim_{x\rightarrow x_0} sin(\frac{1}{x})= sin(\lim_{x\rightarrow x_0}\frac{1}{x})[/tex]
    That inner limit exist for all x0 except 0. The limit as x approaches 0.1 is just sin(1/0.1)= sin(10).

    If x is close enough to 0.1, for example if 1/(0.1+ [itex]\pi[/itex]/2)< x< 1/(0.1- [itex]\pi[/itex]/2) on your graph (x, sin(1/x) ) is on ONE of the many oscillations shown on your graph and the others don't matter.
     
  4. Oct 23, 2008 #3
    hey thank you very much. infact even i realized the mistake in my question after i asked it :). But another question is the way you took the limit inside sin is possible for all continuous curves right?
     
  5. Oct 23, 2008 #4

    HallsofIvy

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    Yes, that's pretty much the definition of continuous!
     
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