Results 1 to 7 of 7

Thread: Hard Limit

  1. #1
    Member
    Joined
    Feb 2008
    Posts
    79

    Hard Limit

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

    $\displaystyle \lim_{(x,y)\to (0,2)}{\frac{6x^3\sqrt{y-2}}{2x^4 + y^2 - 4y + 4}}$
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Senior Member Peritus's Avatar
    Joined
    Nov 2007
    Posts
    397
    $\displaystyle \mathop {\lim }\limits_{(x,y) \to (0,2)} \frac{{6x^3 \sqrt {y - 2} }}
    {{2x^4 + \left( {y - 2} \right)^2 }}
    $

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


    $\displaystyle y = x^2 + 2
    $


    $\displaystyle \mathop {\lim }\limits_{(x,y) \to (0,2)} \frac{{6x^4 }}
    {{3x^4 }} = 2
    $

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


    $\displaystyle y = x^3 + 2$


    $\displaystyle
    \mathop {\lim }\limits_{(x,y) \to (0,2)} \frac{{6x^{4\frac{1}
    {2}} }}
    {{2x^4 + x^6 }} = \mathop {\lim }\limits_{(x,y) \to (0,2)} \frac{{6x^{\frac{1}
    {2}} }}
    {{2 + x^2 }} = 0$

    Since taking different paths toward the same point yields different values for the limit, the limit does not exist.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Feb 2008
    Posts
    79
    Quote Originally Posted by Peritus View Post
    $\displaystyle \mathop {\lim }\limits_{(x,y) \to (0,2)} \frac{{6x^3 \sqrt {y - 2} }}
    {{2x^4 + \left( {y - 2} \right)^2 }}
    $

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


    $\displaystyle y = x^2 + 2
    $


    $\displaystyle \mathop {\lim }\limits_{(x,y) \to (0,2)} \frac{{6x^4 }}
    {{3x^4 }} = 2
    $

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


    $\displaystyle y = x^3 + 2$


    $\displaystyle
    \mathop {\lim }\limits_{(x,y) \to (0,2)} \frac{{6x^{4\frac{1}
    {2}} }}
    {{2x^4 + x^6 }} = \mathop {\lim }\limits_{(x,y) \to (0,2)} \frac{{6x^{\frac{1}
    {2}} }}
    {{2 + x^2 }} = 0$

    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:

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

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

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

    EDIT: when plugging in x^3 + 2, I get (6sign(x))/(x^2+2) .. no clue what the "sign" is.
    Last edited by topsquark; Mar 1st 2008 at 04:55 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,797
    Thanks
    2828
    Awards
    1
    Actually you are correct to question that reply.
    It should be $\displaystyle 6x^3 \sqrt {x^2 } = 6x^3 \left| x \right|$.
    As for 'sign' it is usually written $\displaystyle \left\{ {{\mathop{\rm sgn}} (x) = \begin{array}{ll}
    {1\,,\,x > 0} \\
    {0\,,\,x = 0} \\
    { - 1\,,\,x < 0} \\
    \end{array}} \right.$

    That fits with the absolute value.
    Nonetheless, the idea given in the first reply is correct.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Senior Member Peritus's Avatar
    Joined
    Nov 2007
    Posts
    397
    a little correction: after the first substitution we get:


    $\displaystyle 2\frac{{\left| x \right|}}
    {x}$

    when x approaches 0 the limit of this function is undefined.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Feb 2008
    Posts
    79
    Quote Originally Posted by Peritus View Post
    a little correction: after the first substitution we get:


    $\displaystyle 2\frac{{\left| x \right|}}
    {x}$

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

    For $\displaystyle y = x^2 +2$, we get:

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

    Then, for $\displaystyle y=x^3 + 2$, we get $\displaystyle \frac{6(x^3)^{3/2}}{x^6 + 2x^4}$ and according to my calculator, as $\displaystyle x\to 0$, I'm unable to calculate the limit... (but since we already got one undefined value, the limit doesn't exist?)
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Senior Member Peritus's Avatar
    Joined
    Nov 2007
    Posts
    397
    now as I've already shown you divide the expression by x^4:


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

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


    $\displaystyle \frac{{6 \cdot 0}}
    {{0 + 2}} = 0$




    Function calculator
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Very Hard Limit
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Apr 19th 2011, 10:09 PM
  2. hard limit.
    Posted in the Pre-Calculus Forum
    Replies: 8
    Last Post: Feb 22nd 2011, 05:58 AM
  3. hard limit
    Posted in the Calculus Forum
    Replies: 4
    Last Post: Oct 23rd 2009, 03:15 AM
  4. Another hard limit
    Posted in the Calculus Forum
    Replies: 11
    Last Post: Dec 7th 2007, 08:06 AM
  5. Hard Limit
    Posted in the Calculus Forum
    Replies: 6
    Last Post: Nov 4th 2007, 08:56 PM

Search Tags


/mathhelpforum @mathhelpforum