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Squeeze theorem

  1. Apr 3, 2014 #1
    Suppose I want to find the limit of a function g(x) as lim x-> 0

    f(x) =< g(x) =<h(x)

    My problem is selecting an upper and lower bound.

    Suppose I have to find lim x->0 sin(x)/x

    I know that sin(x) have a lower bound of -1 and upper bound of 1.

    How should I further take this to solve the problem?
     
  2. jcsd
  3. Apr 3, 2014 #2

    Ray Vickson

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    Using your crude bounds, you can conclude that the limit of sin(x)/x lies between -∞ and +∞, if the limit exists at all. I think you will agree that this information is not very useful. Actually, you can improve it to lying between 0 and +∞, but that is not a whole lot better.

    The limit is 100% standard, and appears in every calculus textbook and in many on-line calculus notes/tutorials.
     
  4. Apr 3, 2014 #3

    pasmith

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    In practise one requires that [itex]\lim_{x \to 0} f(x) = \lim_{x \to 0} h(x)[/itex] for the squeeze therorem to be any use. However this problem is best solved by observing that
    [tex]
    \lim_{x \to 0} \frac{\sin(x)}{x}
    [/tex]
    is, if it exists, equal by definition to [itex]\sin'(0)[/itex]. If you already know by other methods (such as by formally defining [itex]\sin x[/itex] to be the solution of [itex]f'' + f = 0[/itex] which satisfies [itex]f(0) = 0[/itex] and [itex]f'(0) = 1[/itex]) that [itex]\sin'(0) = 1[/itex] then you're done, but otherwise you will have to proceed by looking at a suitable right-angled triangle and claiming that [itex]\sin x \approx x[/itex] for [itex]|x|[/itex] sufficiently small.
     
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