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Does the lim x→0 of (x^4)(sin1/x) exist?

  1. Dec 18, 2011 #1
    1. The problem statement, all variables and given/known data
    I need to solve this limit.
    lim (x^4)(sin1/x)
    x→0

    2. Relevant equations



    3. The attempt at a solution

    Since as 1/x → 0+, the limit is +∞ but when 1/x → 0-, the limit is -∞. Since these two-sided limits don't match I thought the limit could not be calculated. apparently this is incorrect. any thoughts?
     
  2. jcsd
  3. Dec 18, 2011 #2
    Yes it does exist. Basically you have to rearrange the function (I know I'm not using great math vocab here).

    You take x^4*sin(1/x) and make it x^4 / (1/sin(1/x)). you then have 0/1...which the limit is 0.

    Correct me if I am way off but that is how I would do it.
     
    Last edited: Dec 18, 2011
  4. Dec 18, 2011 #3
    The limit as x approaches 0 of sin(1/x) does not exist, so you can't use the usual theorems about products (or quotients) of limits.

    But since [itex]|\sin x| \leq 1[/itex],
    [tex]|x^4 \sin(1/x)| = |x^4| \cdot |\sin(1/x)| \leq |x^4| \cdot 1[/tex]
    so the limit does exist and is zero.
     
  5. Dec 18, 2011 #4
    Yep, thats right.
     
  6. Dec 18, 2011 #5
    awesome. thanks guys!
     
  7. Dec 18, 2011 #6

    D H

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    The problem isn't asking about 1/x. It's asking about sin(1/x), and in particular, the it is asking about the product of x4 with the sine of 1/x. You are correct in the sense that sin(1/x) is undefined as x→0. It is however bounded: [itex]-1\le \sin(1/x) \le 1[/itex] for all x. Thus [itex]-x^4\le x^4\sin(1/x)≤x^4[/itex]. What does the squeeze theorem say about this as x→0?
     
  8. Dec 18, 2011 #7
    that since the limits on both of those sides equals 0 then the limit must also equal 0. that is extremely helpful. Thank you!
     
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