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Limit questions

  1. Apr 18, 2007 #1
    i added a file in wich i tried to solved them

    in every limit i solved plz correct me where i get it wrong.

    because some how my answer differs alot from the book answer

    maybe it some basic knowledge that i lack

    plz help
     

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  2. jcsd
  3. Apr 18, 2007 #2
    Could not read your first problem.
    What is the meaning of lin(x)/x ???
     
  4. Apr 18, 2007 #3
    its S
    sinx/x

    lim (sinx)/x =1
    x>0
     
  5. Apr 18, 2007 #4
    Then the first is easy.
    It is of the form lim (f(x))^(sinx/x),
    and lim f(x) is easyli obtained.
     
  6. Apr 18, 2007 #5
    so did i get it right??
    what about the rest??
     
  7. Apr 19, 2007 #6

    VietDao29

    User Avatar
    Homework Helper

    Nope, nope, it's incorrect.
    You should note that the limit for e is:
    [tex]\lim_{x \rightarrow \infty} \left( 1 + \frac{1}{x} \right) ^ x[/tex]

    Or

    [tex]\lim_{x \rightarrow 0} \left( 1 + x\right) ^ \frac{1}{x}[/tex]

    All the two are of the Indeterminate Form [tex]1 ^ \infty[/tex], whereas in your problem, it's not any Indeterminate Forms.

    --------------------

    [tex]\lim_{x \rightarrow 0} \left( \frac{x ^ 2 - 2x + 3}{x ^ 2 - 3x + 2} \right) ^ \frac{\sin x}{x} = \left( \frac{0 ^ 2 - 2 \times 0 + 3}{0 ^ 2 - 3 \times 0 + 2} \right) ^ 1 = \frac{3}{2}[/tex]

    --------------------

    The seconds problem you did it all correct, except for the last part, which should read:
    [tex]= \fbox{\ln} e = 1[/tex]

    --------------------

    The third problem, you've differentiated it incorrectly. :frown:

    [tex](\arctan (2x))' = \frac{2}{1 + 4x ^ 2}[/tex]
    [tex](\sin (3x))' = 3 \cos (3x)[/tex]

    Ok, now, can you complete the three problems on your own? :)
     
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