Results 1 to 9 of 9

Math Help - Quick question related to limits

  1. #1
    Junior Member
    Joined
    Apr 2012
    From
    australia
    Posts
    30

    Quick question related to limits

    say we want the limit of x -> -4 for f(x) = (2x^2 + 6x + 8) / (x^2 + 4)

    is it not possible to simplify the numerator? I am guessing I did the wrong thing when I simplified the numerator to 2((x + 2)(x + 4))?

    So I have some questions.

    1. Is it possible to simplify the numerator?
    2. If the numerator was actually 2((x + 2)(x + 4))... would the limit of x -> -4 be 0.5?
    3. Is the answer to the original question infinity?

    Thanks guys.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,535
    Thanks
    778

    Re: Quick question related to limits

    Quote Originally Posted by SplashDamage View Post
    say we want the limit of x -> -4 for f(x) = (2x^2 + 6x + 8) / (x^2 + 4)

    is it not possible to simplify the numerator? I am guessing I did the wrong thing when I simplified the numerator to 2((x + 2)(x + 4))?
    2((x + 2)(x + 4)) = 2x + 12x + 16, not 2x + 6x + 8. The latter polynomial has no real roots, so it cannot be factored over reals.

    Quote Originally Posted by SplashDamage View Post
    1. Is it possible to simplify the numerator?
    No.
    Quote Originally Posted by SplashDamage View Post
    2. If the numerator was actually 2((x + 2)(x + 4))... would the limit of x -> -4 be 0.5?
    The limit would be 0 because x - 4 = 0 when x = -4.
    Quote Originally Posted by SplashDamage View Post
    3. Is the answer to the original question infinity?
    No, it's 4/5. The denominator is never zero, so the function is continuous for all x, and its limit is its value.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Apr 2012
    From
    australia
    Posts
    30

    Re: Quick question related to limits

    Ok thanks. I see what I did wrong now. Also:

    Basically I have no idea what this is trying to tell me. This is from the online lecture notes my lecturer put up. I have no idea what is going on, and no idea what h is.



    1. Can someone please explain what is going on in steps, really simply.
    2. what is h?

    I know the basic idea is to make the two points closer and closer together, and then get the gradient of those two points when they are almost identical... But as you can see I don't get how to do it and what this is telling me.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,535
    Thanks
    778

    Re: Quick question related to limits

    Quote Originally Posted by SplashDamage View Post
    Basically I have no idea what this is trying to tell me. This is from the online lecture notes my lecturer put up. I have no idea what is going on, and no idea what h is.
    There is not much to explain. P and Q are points on the parabola. The number h is the difference between the x-coordinates of P and Q. The slope of the line PQ, by definition, is the difference of the y-coordinates divided by the difference of the x-coordinates. Let's denote f(x) = x. It is assumed that P has coordinates (1, f(1)) = (1, 1), so Q has coordinates (1 + h, f(1 + h)) = (1 + h, (1 + h)). Therefore, the slope is (f(1 + h) - f(1)) / h. As h tends to zero, the line PQ becomes the tangent to the parabola at P. (This last statement is informal at this point.)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Apr 2012
    From
    australia
    Posts
    30

    Re: Quick question related to limits

    Ah ok, so h is just the difference between x1 and x2?

    Makes sense to me. Thanks man.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,547
    Thanks
    1418

    Re: Quick question related to limits

    Quote Originally Posted by SplashDamage View Post
    say we want the limit of x -> -4 for f(x) = (2x^2 + 6x + 8) / (x^2 + 4)

    is it not possible to simplify the numerator? I am guessing I did the wrong thing when I simplified the numerator to 2((x + 2)(x + 4))?

    So I have some questions.

    1. Is it possible to simplify the numerator?
    2. If the numerator was actually 2((x + 2)(x + 4))... would the limit of x -> -4 be 0.5?
    3. Is the answer to the original question infinity?

    Thanks guys.
    Since the function is defined and continuous for all x, you can evaluate this limit by substituting x = -4.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,693
    Thanks
    1466

    Re: Quick question related to limits

    Quote Originally Posted by SplashDamage View Post
    say we want the limit of x -> -4 for f(x) = (2x^2 + 6x + 8) / (x^2 + 4)

    is it not possible to simplify the numerator? I am guessing I did the wrong thing when I simplified the numerator to 2((x + 2)(x + 4))?
    Here's a quick test: if x= -4, [itex]2x^2+ 6x+ 8= 2(-4)^2+ 6(-4)+ 8= 2(16)- 24+ 8= 32- 24+ 8= 8+ 8= 16. The fact that the value is not 0 tells you that x+ 4 is NOT a factor of the numerator. More to the point, since the denominator goes to 0 while the numerator does NOT shoulde tell you immediately that the limit does not exist.

    IF the numerator had been [itex]2x^2+ 6x- 8[/itex], then with x= -4 we would have 2(16)- 24- 8= 8- 8= 0 and then we would know that x+ 4 is a factor: [itex]2x^2+ 6x- 8= 2(x^2+ 3x- 4)= 2(x+ 4)(x- 1)[/itex].

    So I have some questions.

    1. Is it possible to simplify the numerator?
    2. If the numerator was actually 2((x + 2)(x + 4))... would the limit of x -> -4 be 0.5?
    3. Is the answer to the original question infinity?

    Thanks guys.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor
    Joined
    Oct 2009
    Posts
    5,535
    Thanks
    778

    Re: Quick question related to limits

    Quote Originally Posted by HallsofIvy View Post
    More to the point, since the denominator goes to 0 while the numerator does NOT shoulde tell you immediately that the limit does not exist.
    The denominator does not go to zero, it's >= 4 for all x. If the denominator were x + 4 instead of x^2 + 4, then it would go to zero when x -> -4 .
    Follow Math Help Forum on Facebook and Google+

  9. #9
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,389
    Thanks
    757

    Re: Quick question related to limits

    Quote Originally Posted by SplashDamage View Post
    Ok thanks. I see what I did wrong now. Also:

    Basically I have no idea what this is trying to tell me. This is from the online lecture notes my lecturer put up. I have no idea what is going on, and no idea what h is.



    1. Can someone please explain what is going on in steps, really simply.
    2. what is h?

    I know the basic idea is to make the two points closer and closer together, and then get the gradient of those two points when they are almost identical... But as you can see I don't get how to do it and what this is telling me.

    as you may (or perhaps may not) recall, the slope of the line going through two points (x1,y1) and (x2,y2) is:

    (y2 - y1)/(x2 - x1).

    if our two points lie on the graph of a function f, such as f(x) = x2, then:

    y2 = f(x2) = (x2)2
    y1 = f(x1) = (x1)2.

    suppose further, that we want to know what is happening "near x = a", so we take x1 = a, and let x2 = a+h (where h is how far away x2 is from x1).

    so when x2 gets near x1, then a+h gets near a, that is: h must be getting near 0.

    so now our slope formula becomes:

    (y2 - y1)/(x2 - x1) = (f(x2) - f(x1))/(x2 - x1)

    = (f(a+h) - f(a))/(a+h - a) = (f(a+h) - f(a))/h

    now it looks like if we let h EQUAL 0, we have (f(a) - f(a))/0 = 0/0, which is bad, because 0/0 makes no sense.

    but watch what happens if we just let h be NEAR 0, for f(x) = x2:

    (f(a+h) - f(a))/h = ((a+h)2 - a2)/h = (a2 + 2ah + h2 - a2)/h

    = (2ah + h2)/h = (2ah)/h + h2/h = 2a + h.

    so when h is NEAR 0, the slope of our line is NEAR 2a.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. age related question
    Posted in the Algebra Forum
    Replies: 2
    Last Post: December 11th 2011, 07:26 AM
  2. Limits question... quick
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 11th 2010, 12:44 PM
  3. Quick help in Limits
    Posted in the Calculus Forum
    Replies: 8
    Last Post: August 25th 2008, 12:51 PM
  4. Quick question. related rates
    Posted in the Calculus Forum
    Replies: 2
    Last Post: December 3rd 2006, 01:54 PM
  5. Quick's quick question
    Posted in the Number Theory Forum
    Replies: 22
    Last Post: July 9th 2006, 04:38 PM

Search Tags


/mathhelpforum @mathhelpforum