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Lim x -> inf. of sinx/x

  1. Sep 18, 2012 #1
    1. The problem statement, all variables and given/known data
    [tex]\lim_{x \to \infty } \frac{\sin x}{x}[/tex]

    2. Relevant equations

    3. The attempt at a solution
    [tex]\lim_{x \to \infty } \frac{\sin x}{x}[/tex]
    Substitute x = 1/u
    [tex]\lim_{\frac{1}{u} \to 0 } \frac{\sin \frac{1}{u}}{\frac{1}{u}} = 1[/tex]

    I know this is incorrect because intuitively I know that sinx will oscillate between 0 and 1, but the denominator will grow larger infinitely, therefore should theoretically approach 0 but that's just my thinking and no hard evidence.
  2. jcsd
  3. Sep 18, 2012 #2


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    Translate what you correctly know into an argument using the squeeze theorem.
  4. Sep 18, 2012 #3
    We haven't learned the squeeze theorem, so should I just take sinx/x as x-> inf as 0 for granted and just memorize it like I do with sinx/x as x-> 0? (We were given sinx/x as x-> 0 to be 1, but were told to just memorize it)
  5. Sep 18, 2012 #4


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    No, there's no need to memorize it. Remember -1<=sin(x)<=1, so -1/x<=sin(x)/x<=1/x. Even if you don't have the theorem that should make it pretty obvious that the limit is 0.
  6. Sep 18, 2012 #5


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    That doesn't look right, if you're going to plug 1/u in for x then everything else needs to stay the same. As you have it there, you've changed the limit from [itex]x\to\infty[/itex] to [itex]\frac{1}{u} \to 0[/itex] when it should be [itex]\frac{1}{u} \to \infty[/itex]
    But regardless, this won't help you.

    Well then I guess you can just memorize it, but the squeeze theorem isn't very hard to understand really, especially for this example.

    It's true that [tex]-1\leq \sin(x)\leq 1[/tex] right?
    So if we divide all through by x, we get [tex]\frac{-1}{x}\leq \frac{\sin(x)}{x}\leq \frac{1}{x}[/tex]
    And now we know that the limit as [itex]x\to\infty[/itex] for both [itex]-1/x[/itex] and [itex]1/x[/itex] is equal to zero, and since [itex]\sin(x)/x[/itex] is in between those, its limit must also be zero.
  7. Sep 18, 2012 #6
    Wow, thank you. That helps a ton, so now I know how to explicitly state this
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