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Homework Help: Weird limit question

  1. Oct 9, 2008 #1
    1. The problem statement, all variables and given/known data

    f(x+y)= f(x) + f(y) + y*x^2 +x*y^2 Given: lim of f(x)/x where x approaches 0 is 1
    Find : 1) f(0) 2) f ' (0) 3) f ' (x)
    2. Relevant equations

    3. The attempt at a solution
    countless hours with other coursemates that lead to nothing but a looping headache.
  2. jcsd
  3. Oct 9, 2008 #2


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    Homework Helper

    Note that

    f(x+y) = f(x) + f(y) + x^2 y + x y^2

    means that (using [tex] x = y = 0 [/tex]

    f(0) = f(0) + f(0) + 0

    What does this tell you about [tex] f(0) [/tex]? Also remember the definition:

    f'(0) = \lim_{h \to 0} \frac{f(0+h) - f(0)}{(0+h) - 0} = \lim_{h \to 0} \frac{f(h)-f(0)} h

    This second point relates to one thing you are given
  4. Oct 9, 2008 #3
    Yep, that be true. It's the 3rd part of the question that i can't still solve. I've posted the first two parts of the question as a guide to solving the third part. So any ideas on the 3rd part? Thanks for making the second part ans more organize. My solution was a little more messy.
  5. Oct 10, 2008 #4
    but x+y=0, x can be -y. that is, x and y may not equal to zero?
    so f(0)=f(x)+f(-x)?
  6. Oct 10, 2008 #5


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    [tex] f'(x)= \lim_{h \to 0} \frac{f(x+h)-f(x)}{(x+h)-x}[/tex]

    But given that [itex]f(x+y)=f(x)+f(y)+x^2y+xy^2[/itex] , what is [itex]f(x+h)[/itex] ?
  7. Oct 13, 2008 #6
    Wow, true, I didn't see it that way.Thanks a lot. LOL
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