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Did i solved this limit correctly

  1. Feb 11, 2009 #1
    [tex]
    lim_{x->\infty}\frac{3x-cosx}{4x+sinx}=lim_{x->\infty}\frac{\frac{3x-cosx}{x}}{\frac{4x+sinx}{x}}=lim_{x->\infty}\frac{3-0}{4-0}=0.75
    [/tex]

    i thought bounded/infinity=0


    ??
     
  2. jcsd
  3. Feb 12, 2009 #2

    Mark44

    Staff: Mentor

    You have the correct limit, but you should remove the limit operator from the expression below, since you have already "passed to the limit" in the previous expression:
    [tex]lim_{x->\infty}\frac{3-0}{4-0}=0.75 [/tex]
     
  4. Feb 12, 2009 #3
    thanks
     
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