Results 1 to 11 of 11
Like Tree5Thanks
  • 1 Post By Jonroberts74
  • 1 Post By ManuelSG
  • 2 Post By Plato
  • 1 Post By Jonroberts74

Math Help - Calculate the limit if it exists.

  1. #1
    Member
    Joined
    May 2011
    Posts
    133
    Thanks
    2

    Calculate the limit if it exists.

    I wasn't sure if to post this in the Calculus forum or in the Algebra forum. My question has more to do with the algebra than it does with the calculus part however since it is a limit question I thought it would be better to post to the Calculus forum.

    Calculate the limit if it exists.-calculatethelimit.png
    Calculate the limit if it exists.-calculatethelimitsolution.png
    Calculate the limit if it exists.-calculatethelimitscreenshot.png

    It looks to me by looking at the graph that as x->infinity that y-> -infinity. Problem f) seems to be solved positive infinity? I'm still trying to figure out the algebra that I highlighted I don't see yet how those are equivalent. Also I was wondering am I supposed to know the answer to f) without looking at its graph or how if that is possible?

    Thanks in advance...
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Member
    Joined
    Sep 2013
    From
    Portland
    Posts
    225
    Thanks
    7

    Re: Calculate the limit if it exists.

    type out your work and then I can help you find the error

    on a side note - learning to use LaTeX will greatly aid in typing out your work. I find that because it takes more time than writing it out on paper you get a better understanding of a problem you are having problems with. It is fairly easy to learn.
    Last edited by Jonroberts74; October 29th 2013 at 03:10 PM.
    Thanks from sepoto
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jun 2013
    From
    Chile
    Posts
    4
    Thanks
    1

    Re: Calculate the limit if it exists.

    You have to multiply the whole fraction by \frac { \frac { 1 }{ x }  }{ \frac { 1 }{ x }  } wich is secretly 1 so you are not changing the function.
    \frac { 4{ x }^{ 2 } }{ x-2 } \ast \frac { \frac { 1 }{ x }  }{ \frac { 1 }{ x }  } =\frac { \frac { 4{ x }^{ 2 } }{ x }  }{ \frac { x-2 }{ x }  } =\frac { 4x }{ \frac { x }{ x } -\frac { 2 }{ x }  } =\frac { 4x }{ 1-\frac { 2 }{ x }  } so we say both fractions are equivalent and the second form is useful to find the limit.

    I hope this is useful

    PD: You could also use L'H˘pital Rule to find this limit
    Last edited by ManuelSG; October 29th 2013 at 03:09 PM.
    Thanks from sepoto
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,403
    Thanks
    1486
    Awards
    1

    Re: Calculate the limit if it exists.

    Quote Originally Posted by ManuelSG View Post
    You have to multiply the whole fraction by \frac { \frac { 1 }{ x }  }{ \frac { 1 }{ x }  } wich is secretly 1 so you are not changing the function.
    Why do you say that? There is nothing secret about that.

    But why not just say "divide the numerator and the denominator by x"?
    Last edited by Plato; October 29th 2013 at 03:53 PM.
    Thanks from sepoto and ManuelSG
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Sep 2013
    From
    Portland
    Posts
    225
    Thanks
    7

    Re: Calculate the limit if it exists.

    Oh another trick in your book is looking at the degree of the dominant term in the denominator and numerator.

    And,

    If the degree in the dominant term of both the denominator and numerator are equal you'll end up with a horizontal asymptote at y=0

    or

    If the degree of the dominant term in the denominator is greater the the degree of the dominant term in the numerator you will have a horizontal asymptote

    at y = \frac{coefficient\, of\, the\, dominant\, term\, in\, the\, numerator}{coefficient\, of\, the\, dominant\, term\, in\, the\, denominator}

    but in yours the greater degree of a dominant term is in your numerator thus f(x) \rightarrow \pm \infty \,as\,x\rightarrow \pm \infty

    And then you can test the behavior of the function by creating the quotient \frac{dominant\, term\, in\, the\, numerator}{dominant\, term\, in\, the\, denominator}

    and see how it behaves when x gets large and negative and large and positive


    Also,

    it might help you to remember

    \lim_{x \ \pm\infty} \frac{1}{x} = 0 so the idea is to express f(x) in terms of \frac{1}{x}

    hope those help!
    Last edited by Jonroberts74; October 29th 2013 at 03:57 PM.
    Thanks from sepoto
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    May 2011
    Posts
    133
    Thanks
    2

    Re: Calculate the limit if it exists.

    To begin with thank you for all your very helpful replies!

    \frac{4x}{1-\frac{2}{x}}

    I see now how the above equation was derived. Also thank you Jonroberts74 for posting:

    \lim_{x\pm\infty}\frac{1}{x}=0
    The above is a concept that I can understand.

    In my solutions manual there is an equation:

    \frac{4x}{1-\frac{2}{x}}\rightarrow \frac{\infty}{1}

    I have to admit I have been looking at it for quite some time and I don't see yet what is meant by:

    {\rightarrow \frac{\infty}{1}}

    Also as I have graphed the equation which is attached to my original post and by looking only at the graph there is no f(x) where x is positive that yields a positive y value so I still don't see how the limit of the function is positive infinity as x goes in the positive direction towards infinity.
    Last edited by sepoto; October 29th 2013 at 04:46 PM.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    May 2011
    Posts
    133
    Thanks
    2

    Re: Calculate the limit if it exists.

    I had to fix some Latex errors. My tag should have been [tex] not [math]
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Sep 2013
    From
    Portland
    Posts
    225
    Thanks
    7

    Re: Calculate the limit if it exists.

    \frac{\infty}{1} = \infty\frac{1}{1} = \infty

    thus as\, f(x) \rightarrow \infty

    also sorry, it should be

    \lim_{x\rightarrow\pm\infty} \frac{1}{x} = 0 and not \lim_{x\pm\infty} \frac{1}{x} = 0 I forgot the \rightarrow
    Last edited by Jonroberts74; October 29th 2013 at 05:01 PM.
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    May 2011
    Posts
    133
    Thanks
    2

    Re: Calculate the limit if it exists.

    \frac{4\infty}{1-\frac{2}{\infty}}\rightarrow \frac{\infty}{1-0}

    So could I say that the equation of the left and the equation on the right are equivalent?
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Sep 2013
    From
    Portland
    Posts
    225
    Thanks
    7

    Re: Calculate the limit if it exists.

    Quote Originally Posted by sepoto View Post
    \frac{4\infty}{1-\frac{2}{\infty}}\rightarrow \frac{\infty}{1-0}

    So could I say that the equation of the left and the equation on the right are equivalent?
    Well this might be getting into some territory that I don't quite know yet but to me it seems like you are now treating infinity as though it is a number or variable which is incorrect.

    Regardless,

    I would state it as "f(x) tends to infinity".
    Last edited by Jonroberts74; October 29th 2013 at 05:12 PM.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    May 2011
    Posts
    133
    Thanks
    2

    Re: Calculate the limit if it exists.

    It looks like I should have zoomed out quite a lot with my graph.Calculate the limit if it exists.-screenshot-4-.png
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Evaluate the limit, if it exists.
    Posted in the Calculus Forum
    Replies: 2
    Last Post: May 11th 2013, 02:07 PM
  2. Replies: 2
    Last Post: February 17th 2013, 09:50 AM
  3. Replies: 16
    Last Post: November 15th 2009, 04:18 PM
  4. Verifying a limit exists
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: March 8th 2009, 03:45 AM
  5. showing that the limit exists
    Posted in the Calculus Forum
    Replies: 11
    Last Post: September 25th 2007, 10:01 AM

Search Tags


/mathhelpforum @mathhelpforum