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Squeeze Therom Problem

  1. Oct 9, 2008 #1
    1. The problem statement, all variables and given/known data

    find the lim of:

    lim X-> -2 (3/4x) cos (4/(x+2))



    3. The attempt at a solution

    The x-> -2 threw me off, because the center changed.

    im having trouble understanding the function of the squeeze therm here.
    i understand that the bound is (-3/4x) and (3/4x). i also understand that the graph of the cos is shifted to the left by 2 units and so as x-> -2 it is centered at -2.

    i was confused when i took the lim of (-3/4x) and got (3/8), and for (3/4x) i got (-3/8).
    since the lim don't equal i am going assuming the lim DNE. But on the other hand when i graph (-3/4x) and (3/4x) they make and X graph centered on the origin. Also the graph of the cos(4/(2+x)) oscillates wildly at -2 but cos(4/(2+x)) is also squeezed at the origin between (-3/4x) and (3/4x).

    so would the lim be zero? or would it be DNE? i really think its zero since the bounds of (-3/4x) and (3/4x) squeeze cos (4/(x+2)), but i don't understand how i am supposed to do it with out a graphing calculator or graphing it by hand to see where the functions are being squeezed.

    Thank you in advance.
     
  2. jcsd
  3. Oct 9, 2008 #2

    Dick

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    Your first guess, DNE, is the correct one. And you are also right that very near x=-2 the function oscillates infinitely between about -3/8 and +3/8 (since 4/(2+x) goes to +/-infinity it goes through an infinite number of cycles of cos). That's not enough of a squeeze to force it to a limit. You don't really need a graphing calculator, to make a rough picture of this in your head.
     
  4. Oct 10, 2008 #3
    What do you mean by its "not enough of a squeeze to force it to a lim"?
     
  5. Oct 10, 2008 #4

    Dick

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    I mean you are only forcing the oscillation to be between -3/8 and 3/8. There's a lot of numbers in that range. The limit doesn't exist.
     
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