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Lim of trig functions. Does it exist?

  1. May 30, 2013 #1
    1. Does the limit exist of the following:

    lim as x→ 1- ((cos^-1(x))/(1-x))



    2. Relevant equations: division_law.gif



    3. The attempt at a solution:

    lim as x→ 1- ((cos^-1(x))/(1-x))
    = lim as x→ 1- (cos^-1(x))/ lim as x→ 1-(1-x)

    Let y = 1-x

    lim as y→0 (cos^-1(1-y)) / lim as y→0 (y)
    = 0/0 therfore limit of the entire function as x→1- is ∞
     
    Last edited by a moderator: May 31, 2013
  2. jcsd
  3. May 30, 2013 #2

    SteamKing

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    Sorry, 0/0 is an indeterminate form.
     
  4. May 30, 2013 #3
    So 0/0 = Limit doesnt exist ?
     
  5. May 30, 2013 #4

    SteamKing

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    Not necessarily.

    For example, the limit of sin(x)/x as x approaches 0 is equal to 1.
     
  6. May 31, 2013 #5

    verty

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    Let ##x = cos(y)##, this may help to simplify things.

    If not, I would leave this question for later.
     
  7. May 31, 2013 #6

    Mark44

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    [0/0] is one of several indeterminate forms. The "indeterminate" part means that you can't tell if an expression with this form has a limit, and if it does, what that limit will be.

    Some of the other indeterminate forms are [∞/∞], [∞ - ∞], and [1].

    All of the limits below are of the [0/0] indeterminate form:
    $$ 1. \lim_{x \to 0}\frac{x^2}{x}$$
    $$ 2. \lim_{x \to 0}\frac{x}{x^2}$$
    $$ 3. \lim_{x \to 0}\frac{x}{x}$$
    In #1, the limit is 0; in #2, the limit doesn't exist; in #3, the limit is 1.
     
  8. May 31, 2013 #7

    eumyang

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    I think you switched #1 and #2. In #1, the limit doesn't exist, and in #2, the limit is 0.
     
  9. May 31, 2013 #8

    Mark44

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    I don't think so. Factor x/x out of each and see what you get.
     
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