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Math Help - Hard Limit

  1. #1
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    Hard Limit

    Find the following limit, or show that the limit doesn't exist.

    \lim_{(x,y)\to (0,2)}{\frac{6x^3\sqrt{y-2}}{2x^4 + y^2 - 4y + 4}}
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  2. #2
    Senior Member Peritus's Avatar
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    \mathop {\lim }\limits_{(x,y) \to (0,2)} \frac{{6x^3 \sqrt {y - 2} }}<br />
{{2x^4  + \left( {y - 2} \right)^2 }}<br />

    lets look what's the limit when it is approached along the curve:


    y = x^2  + 2<br />


    \mathop {\lim }\limits_{(x,y) \to (0,2)} \frac{{6x^4 }}<br />
{{3x^4 }} = 2<br />

    now lets look what's the limit when it is approached along the curve:


    y = x^3  + 2


    <br />
\mathop {\lim }\limits_{(x,y) \to (0,2)} \frac{{6x^{4\frac{1}<br />
{2}} }}<br />
{{2x^4  + x^6 }} = \mathop {\lim }\limits_{(x,y) \to (0,2)} \frac{{6x^{\frac{1}<br />
{2}} }}<br />
{{2 + x^2 }} = 0

    Since taking different paths toward the same point yields different values for the limit, the limit does not exist.
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  3. #3
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    Quote Originally Posted by Peritus View Post
    \mathop {\lim }\limits_{(x,y) \to (0,2)} \frac{{6x^3 \sqrt {y - 2} }}<br />
{{2x^4  + \left( {y - 2} \right)^2 }}<br />

    lets look what's the limit when it is approached along the curve:


    y = x^2  + 2<br />


    \mathop {\lim }\limits_{(x,y) \to (0,2)} \frac{{6x^4 }}<br />
{{3x^4 }} = 2<br />

    now lets look what's the limit when it is approached along the curve:


    y = x^3  + 2


    <br />
\mathop {\lim }\limits_{(x,y) \to (0,2)} \frac{{6x^{4\frac{1}<br />
{2}} }}<br />
{{2x^4  + x^6 }} = \mathop {\lim }\limits_{(x,y) \to (0,2)} \frac{{6x^{\frac{1}<br />
{2}} }}<br />
{{2 + x^2 }} = 0

    Since taking different paths toward the same point yields different values for the limit, the limit does not exist.
    When plugging x^2 + 2 in for y, we get:

    \frac{6x^3\sqrt{x^2 + 2 -2}}{2x^4+(x^2+2-2)^2}; how's that equal to \frac{6x^4}{3x^4} .

    And then when plugging in x^3 + 2 for y, we get:

    \frac{18}{x^3 + 2x}. As x \to 2, that limit is \frac{3}{2}..

    EDIT: when plugging in x^3 + 2, I get (6sign(x))/(x^2+2) .. no clue what the "sign" is.
    Last edited by topsquark; March 1st 2008 at 04:55 PM.
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  4. #4
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    Actually you are correct to question that reply.
    It should be 6x^3 \sqrt {x^2 }  = 6x^3 \left| x \right|.
    As for 'sign' it is usually written \left\{ {{\mathop{\rm sgn}} (x) = \begin{array}{ll}<br />
   {1\,,\,x > 0}  \\<br />
   {0\,,\,x = 0}  \\<br />
   { - 1\,,\,x < 0}  \\<br />
\end{array}} \right.

    That fits with the absolute value.
    Nonetheless, the idea given in the first reply is correct.
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  5. #5
    Senior Member Peritus's Avatar
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    a little correction: after the first substitution we get:


    2\frac{{\left| x \right|}}<br />
{x}

    when x approaches 0 the limit of this function is undefined.
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  6. #6
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    Quote Originally Posted by Peritus View Post
    a little correction: after the first substitution we get:


    2\frac{{\left| x \right|}}<br />
{x}

    when x approaches 0 the limit of this function is undefined.
    Ok let me know if I'm wrong.

    For y = x^2 +2, we get:

    \frac{6x|x^3|}{3x^4} and as x\to 0, that is undefined.

    Then, for y=x^3 + 2, we get \frac{6(x^3)^{3/2}}{x^6 + 2x^4} and according to my calculator, as x\to 0, I'm unable to calculate the limit... (but since we already got one undefined value, the limit doesn't exist?)
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  7. #7
    Senior Member Peritus's Avatar
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    now as I've already shown you divide the expression by x^4:


    \frac{{6x^{{\raise0.7ex\hbox{$1$} \!\mathord{\left/<br />
 {\vphantom {1 2}}\right.\kern-\nulldelimiterspace}<br />
\!\lower0.7ex\hbox{$2$}}} }}<br />
{{x^2  + 2}}

    as you can clearly see when x approaches 0 we get:


    \frac{{6 \cdot 0}}<br />
{{0 + 2}} = 0




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