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Math Help - Help in finding the Limits..

  1. #1
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    Help in finding the Limits..

    I need help determining the limits of the following problems.
    Determine the following limits(if the limit exists otherwise explain why it does not) without using the L'Hospital's rule...

    1) Lim [sqrt(4+x) - 2)] / [x]
    x->0


    2) Lim [ x^2 - 1] / [|x-1|]
    x->1


    3) Lim [sqrt(x) . cos (1/x) ] ......using L'Hospital's rule...
    x->0
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Vedicmaths View Post
    2) Lim [ x^2 - 1] / [|x-1|]
    x->1
    \lim_{x \to 1} \frac{x^2 - 1}{|x - 1|}

    As x approaches 1 from the left:
    \lim_{x \to 1^-} \frac{x^2 - 1}{-(x - 1)} = \lim_{x \to 1^-} -(x + 1) = -2

    As x approaches 1 from the right:
    \lim_{x \to 1^+} \frac{x^2 - 1}{x - 1} = \lim_{x \to 1^+} (x + 1) = 2

    Since these two are not the same, the limit does not exist.

    -Dan
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  3. #3
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    Quote Originally Posted by topsquark View Post

    = \lim_{x \to 0} \left ( \sqrt{\frac{4}{x^2} + \frac{1}{x}} - \frac{2}{x} \right )

    It should be clear at this point that the limit is 0.
    At this point?
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Vedicmaths View Post
    3) Lim [sqrt(x) . cos (1/x) ] ......using L'Hospital's rule...
    x->0
    I really don't see how to apply L'Hopital's rule here:
    \lim_{x \to 0} \sqrt{x}~cos \left ( \frac{1}{x} \right )

    Let y = \frac{1}{x}. Then your limit is
    \lim_{x \to 0} \sqrt{x}~cos \left ( \frac{1}{x} \right ) = \lim_{y \to \infty} \frac{cos(y)}{\sqrt{y}}

    This is not an indeterminate form: cos(y) never reaches a limit, but is always between -1 and 1. \sqrt{y}, on the other hand, is unbounded and positive. Thus
    \lim_{x \to 0} \sqrt{x}~cos \left ( \frac{1}{x} \right ) = \lim_{y \to \infty} \frac{cos(y)}{\sqrt{y}} = 0

    -Dan
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  5. #5
    Super Member flyingsquirrel's Avatar
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    Hello
    Quote Originally Posted by Vedicmaths View Post
    1) Lim [sqrt(4+x) - 2)] / [x]
    x->0
    Quote Originally Posted by topsquark
    It should be clear at this point that the limit is 0.
    Let f(x)=\sqrt{4+x}

    Notice that \lim_{x\to 0}\frac{\sqrt{4+x}-2}{x}=\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=f'(0)=\frac{1}{2\sqrt{4}}=\frac{1}{4}
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  6. #6
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by flyingsquirrel View Post
    Hello


    Let f(x)=\sqrt{4+x}

    Notice that \lim_{x\to 0}\frac{\sqrt{4+x}-2}{x}=\lim_{x\to 0}\frac{f(x)-f(0)}{x-0}=f'(0)=\frac{1}{2\sqrt{4}}=\frac{1}{4}
    I was thinking the limit was at infinity for some reason. Ugh!

    Thanks for the catch!

    -Dan
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