Results 1 to 10 of 10

Math Help - Help with limits

  1. #1
    Junior Member
    Joined
    Jan 2010
    Posts
    54

    Help with limits

    Hey guys,

    I am wondering if anyone could help with this limit. I'm new to calculus and I need some help with absolute value.




    Thanks
    Follow Math Help Forum on Facebook and Google+

  2. #2
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    Quote Originally Posted by evant8950 View Post
    Hey guys,

    I am wondering if anyone could help with this limit. I'm new to calculus and I need some help with absolute value.




    Thanks
    Firs we solve g(x)=2x^3-x^2\leq0

    2x^2(x-\frac{1}{2})\leq0\Rightarrow{x}\leq\frac{1}{2}

    Now note that g(x)=|2x^3-x^2|=\left\{\begin{array}{cc}-(2x^3-x^2),&\mbox{ if }x\leq\frac{1}{2}\\\\2x^3-x^2,&\mbox{ if }x>\frac{1}{2}\end{array}\right.

    Since we are concerned with the behavior as x\to.05^- (i.e. x<\frac{1}{2}), we will use the first piece of the function g(x)...

    CCan you continue?
    Last edited by VonNemo19; January 23rd 2010 at 06:56 PM.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Jan 2010
    Posts
    54
    When you factored the the problem is 2x^2 set equal to zero? I see how x is less than or equal to than 1/2. Or is it 1/2 because it makes the equation equal to 0?
    Follow Math Help Forum on Facebook and Google+

  4. #4
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    Quote Originally Posted by evant8950 View Post
    When you factored the the problem is 2x^2 set equal to zero? I see how x is less than or equal to than 1/2. Or is it 1/2 because it makes the equation equal to 0?
    All I've done is extended the idea of the definition of absolute value, which is

    |x|=\left\{\begin{array}{cc}-x,&\mbox{ if }x\leq0\\x,&\mbox{ if }x>0\end{array}\right.

    But, the deal is that we've got to know at what point to define our new peice. In your case, we had to solve that inequality to determine we're the stuff inside the absolute value sign would become negative. More precisely, when 2x^3-x^2 would be less than or equal to 0. As it turns out, from solving the inequality, we find that this is true when x\leq\frac{1}{2}

    Did I answer you question?
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Jan 2010
    Posts
    54
    Yes. That answered my question.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    Jan 2010
    Posts
    54
    When I solve this problem I am getting 1/4 for the answer. Is this correct?

    Thanks
    Follow Math Help Forum on Facebook and Google+

  7. #7
    No one in Particular VonNemo19's Avatar
    Joined
    Apr 2009
    From
    Detroit, MI
    Posts
    1,823
    Quote Originally Posted by evant8950 View Post
    Yes. That answered my question.
    EDIT:

    \lim_{x\to.05^-}\frac{2x-1}{-(2x^3-x^2)}=-\lim_{x\to.05^-}\frac{2x-1}{x^2(2x-1)}=-\lim_{x\to.05^-}\frac{1}{x^2}=-\frac{1}{(.05)^2}
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    Jan 2010
    Posts
    54
    Yes. I've gotten that part.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Junior Member
    Joined
    Jan 2010
    Posts
    54
    Thanks! I was close. I appreciate it!
    Follow Math Help Forum on Facebook and Google+

  10. #10
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,038
    Thanks
    1675
    Quote Originally Posted by evant8950 View Post
    Thanks! I was close. I appreciate it!
    Personally, I don't consider "1/4" to be all that close to "-400"!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Using limits to find other limits
    Posted in the Calculus Forum
    Replies: 7
    Last Post: September 18th 2009, 05:34 PM
  2. Function limits and sequence limits
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: April 26th 2009, 01:45 PM
  3. HELP on LIMITS
    Posted in the Calculus Forum
    Replies: 4
    Last Post: September 23rd 2008, 11:17 PM
  4. Limits
    Posted in the Calculus Forum
    Replies: 5
    Last Post: September 21st 2008, 10:52 PM
  5. [SOLVED] [SOLVED] Limits. LIMITS!
    Posted in the Calculus Forum
    Replies: 1
    Last Post: February 25th 2008, 10:41 PM

Search Tags


/mathhelpforum @mathhelpforum