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Limit definition ?

  1. Mar 29, 2008 #1
    1. The problem statement, all variables and given/known data
    Use the limit definition to find the derivative of the function.
    f(x)= 1/(2x-4)


    2. Relevant equations
    f(x+h)-f(x)/h


    3. The attempt at a solution
    ok so first i plugged it all in..

    (1/(2(x+h)-4))-(1/(2x-4)) / h

    from here i was going to do fly-by with the two top fractions but for some reason nothing cancels out...
    so please help this is due for me by monday!! lol
    (any help appreciated)
     
  2. jcsd
  3. Mar 29, 2008 #2
    Expand it, find a common denominator, move on from there. You're on the right track.
     
  4. Mar 29, 2008 #3
    ok so what about this?

    for the top fractions if you do fly by you get:

    (2x+2h-4 - 2x+4)/(2x+2h-4)(2x+4) and then of course all of that divided by h...

    then you have just 2h on the numerater becuz everything else cancels out...
    so:

    (2h)/ (2x+2h-4)(2x+4) all divided by h...

    so then you can do the following...

    2h/(2x+2h-4)(2x+4) • 1/h (reciprical)

    by doing that you can cancel out the h's

    2/(2x+2h-4)(2x+4) •1...

    thats as far as i can go...is this right? if so what do i do next plz
     
  5. Mar 29, 2008 #4
    That is almost right. Check your denominator again, it should not be (2x+2h-4)(2x+4). The rest of your algebra is correct, and you did well to cancel out the h. Now, recall the definition of a derivative. What should the limit be?
     
  6. Mar 29, 2008 #5
    i have no idea cuz when she was teaching this portion i was making up a quiz :/ so thats why im so lost..
    and should the denominator be (2x+2h-4)(2x-4) ? becuz i thought the whole part with
    f(x+h) - f(x) i thought that changed the sign of each part? due to the subtraction sign
     
  7. Mar 29, 2008 #6
    Yes, the denominator should be (2x+2h-4)(2x-4). Even though you are subtracting two fractions, you still need to find a common denominator, which, in this case, is a product of the two denominators.

    I've taken a quick glance at some of your other threads, and you seem to be getting good advice there. This is the same thing. Apply the limit definition of the derivative that you were given, replacing the generic f(x) with your specific function. Then, you need to simplify to a point where you can replace the variable you are taking to a limit with the limiting value and still get a meaningful answer (something other than 0/0), and you will have the derivative.
     
  8. Mar 29, 2008 #7
    yeah i just dont know what the limit should be for this specific problem.. cuz it didnt give me a point
     
  9. Mar 29, 2008 #8
    You don't need a point. The limit as h->0 is for any point, x, on the derivative of the function. It will be in terms of x.
     
  10. Mar 30, 2008 #9
    so 2/(2(0) + 2h +4)(2(0) -4)
    2/(2h+4)(-4)
    2/(-8h -16)

    is that right? can i just leave the h there like that?
     
  11. Mar 30, 2008 #10
    [tex]\lim_{x\rightarrow a}\frac{\frac{1}{2x-4}-\frac{1}{2a-4}}{x-a}=\lim_{x\rightarrow a}\frac{2a-4-2x+4}{(2x-4)(2a-4)(x-a)}=\lim_{x\rightarrow a}\frac{-2(x-a)}{(2x-4)(2a-4)(x-a)}=-2\lim_{x\rightarrow a}\frac{1}{(2x-4)(2a-4)}=\frac{-2}{(2a-4)^{2}}=f'(a), \forall a[/tex]



    P.S. YOu have another post with the exact same question.
     
    Last edited: Mar 30, 2008
  12. Mar 30, 2008 #11
    It's as h->0, not as x->0. Also, 2/ ((2x+2h+4)(2x-4)) is incorrect.
     
  13. Mar 30, 2008 #12
    oops lol
    dont i feel dumb...

    ok so its:
    2/(2x+2(0)-4)(2x-4)
    2/(2x-4)(2x-4)
    2/4x^2-16x+16
    and that would be my answer?
     
  14. Mar 30, 2008 #13
    [tex]\lim_{h\rightarrow 0}\frac{\frac{1}{2x+2h-4}-\frac{1}{2x-4}}{h}[/tex]

    Do it again starting from here, mate. Your answer's not right.
     
    Last edited: Mar 30, 2008
  15. Mar 30, 2008 #14

    Astronuc

    User Avatar

    Staff: Mentor

    Precal_Chris,

    Do as Snazzy suggested and this time write out the numerator paying attention to the order.
     
  16. Mar 30, 2008 #15
    oh ok..so wait when i do fly by... is my numerator supposed to be:

    2x+2h-4-2x-4?

    which would ultimately make it..

    2h-8 then the h's cancel so i have

    -6 in the numerator?
     
  17. Mar 30, 2008 #16

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    No, it is (2x+2h-4)-(2x-4)

    Be careful with your signs!!
     
  18. Mar 30, 2008 #17
    well then wouldnt my first answer be right?
    cuz
    2x+2h-4 -2x +4 would cancel the 2x and the 4's out
    so id be left with 2h..
    then the h would cancel
    and id have
    2/(2x+2h-4)(2x-4)

    and then after putting in the 0's for the h..
    id have

    2/(2x-4)(2x-4)
    so that would be
    2/(2x-4)^2
    or
    2/(4x^2-16x+16)
    right?
     
  19. Mar 30, 2008 #18
    You are still short of a minus sign, check it with post #10.
     
  20. Mar 30, 2008 #19
    in post #10 i didnt understand nething on it..it had a different equation than what i was working with
    but i already had other people tell me that it was right on this ...so when i brought it up later it should still be right?
     
  21. Mar 30, 2008 #20
    oh wait i understand now...
    when i did flyby i did it in the incorrect order...making the numerator wrong...
    instead of:

    (2x+2h-4)-(2x-4)
    it should have been

    (2x-4)-(2x+2h-4)

    then after losing the x's and the 4's i'd have:

    -2h/(2x+2h-4)(2x-4) • 1/h

    then i get rid of the h

    so:
    -2/(2x+2h-4)(2x-4)

    then plug in 0 for h

    -2/(2x+0-4)(2x-4)

    -2/(2x-4)^2

    is that my final answer?
     
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