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Homework Help: Lim sin(1/x) as x->inf

  1. Sep 24, 2010 #1
    1. The problem statement, all variables and given/known data
    lim (x->inf) sin(1/x)

    i have a teacher that seems to think this is equal to 1.. I don't see how this is correct
    as x approaches infinity 1/x aproaches zero... sin 0 = 0 right?
    or is this the wrong way of thinking?


    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 24, 2010 #2

    vela

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    You're right. Perhaps your teacher was thinking of

    [tex]\lim_{x \to 0} \frac{\sin x}{x} = 1[/tex]
     
  4. Sep 24, 2010 #3
    Thanks very much yes maybe he was thinking of that. but here is the actuall email he sent out correcting himself... if anything it confused me more... does what he say hold true?
    "
    The application of the squeeze theorem that we did in class to lim(x->inf)sin(1/x)/x, was correct and gives the answer 0, but I was wrong to say that lim(x->inf)sin(1/x) DNE, it is as some students saw 1. What I intended was to consider the limit: lim(x->0) x sin(1/x). Now lim(x->0)sin(1/x) = DNE because as x->0, sin(1/x) oscillates wildly between -1 an +1, so ones really needs the squeeze theorem here!"
     
  5. Sep 24, 2010 #4
    nop 1/x --->0 so sin(1/x) goes to zero
     
  6. Sep 24, 2010 #5
    well if u know L'hospital rule you can use it on [tex]
    \lim_{x \to 0} \frac{\sin x}{x} = 1
    [/tex]

    so u get [tex]
    \lim_{x \to 0} \frac{\cos x}{1} = 1
    [/tex]
     
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