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Limit x->0 of f(x)/x

  1. Aug 27, 2011 #1
    1. The problem statement, all variables and given/known data
    Given: f(x+y)=f(x) + f(y)+x2y+xy2
    limit[itex]_{x->0}[/itex][itex]\frac{f(x)}{x}[/itex]=1

    2. Relevant equations

    What is f(0)?

    3. The attempt at a solution

    The answer is f(0)=0 because as x->0 then f(x)->0. so f(0)=0

    That is not my attempt. That is the solution. I am stuck from the beginning and don't know where to start. Can someone give me a hint as to how to start ?
     
  2. jcsd
  3. Aug 27, 2011 #2

    gb7nash

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    I'm not sure why a limit existing implies the value of the function. Unless I'm missing something, you don't need the limit at all. Look at the first piece of information. Let x = y = 0, so:

    f(0+0)=f(0) + f(0)+02*0+0*02, so...
     
  4. Aug 28, 2011 #3
    I don't think that would work. what if x=2 and y=-2. and you still don't know what f(x) is and f(y) is.
     
  5. Aug 28, 2011 #4



    f(0)=0

    can be seen clearly when x,y are set to zero
     
  6. Aug 28, 2011 #5
    ok well if 2 people say it is then I will believe that. Thanks.
     
  7. Aug 28, 2011 #6

    vela

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    The original answer you were given is incorrect or, at best, incomplete.

    The statement that the only way the limit [tex]\lim_{x \to 0} \frac{f(x)}{x} = 1[/tex]can exist is if [tex]\lim_{x \to 0} f(x) = 0[/tex]is correct. Why? Because if f(x) didn't approach 0 as x went to 0, the quantity f(x)/x would diverge in the same limit, so the original limit couldn't exist.

    However, saying[tex]\lim_{x \to 0} f(x) = 0[/tex]is not the same as saying f(0)=0. The limit only tells you what the function is doing near the origin. It says absolutely nothing about what the function actually does at the origin. The function could be discontinuous at x=0 or even undefined, and the limit could still exist.

    Now if you can use the given information about f(x) to show it's continuous, then yes, it follows that f(0) = 0 from the limit. A complete solution based on the limit would have to show f(x) is continuous.
     
    Last edited: Aug 28, 2011
  8. Aug 28, 2011 #7
    So are you saying that the problem lacks the information to find out f(0) and that the answer arrived upon, 0, is only partially complete? Should I even consider the first equation for f(x+y) when trying to find f(0), because there are others saying f(0) means that x and y are also zero.
     
  9. Aug 28, 2011 #8

    uart

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    The point Vela makes is that the given limit implies that [itex]f(x) \rightarrow 0[/itex] as [itex]x \rightarrow 0[/itex], but this is not quite the same thing as saying f(0)=0.

    For an example that illustrates the difference, say [itex]f(x) = \frac{1-\cos(x)}{x}[/itex]. Then the limit of f(x) as x goes to zero is certainly zero, but f(0) is undefined.

    The other method that people have tried to point out to you is to replace both x and y with zero in the given definition of f(x+y).

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

    which gives : f(0) = 2 f(0)

    Subtracting f(0) from both sided gives f(0) = 0.
     
  10. Aug 28, 2011 #9

    uart

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    BTW. Just out of interest, you can use the given data to find the first principles derivative of f(x) and hence establish an explicit function for f. You get,

    [tex] f(x) = x + \frac{1}{3}x^3[/tex]
     
  11. Aug 28, 2011 #10
    Thank you uart I understand that explanation very well.

    Side question. because the denominator is x and x->0 wouldn't that make the equation undefined? In this case how would I approach this problem since there is no factor for me to cancel and f(x) is not a factor with numbers or anything I can use to cancel out x. How would I apply the limit to both top and bottom of the f(x)/x as x->0?
     
    Last edited: Aug 28, 2011
  12. Aug 28, 2011 #11
    @ uart
    I calculated f'(x) using the definition of derivative to be x2+1
    integrating to get f(x) would that add a constant C to f(x) or can that be omited?
     
  13. Aug 28, 2011 #12

    uart

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    Yes that's correct. I used the given f(x+y) expression to find lim (f(x+h) - f(x)) / h and do the first principle derivative.

    You get f'(x) = 1 + x^2 and hence f(x) = x + 1/3 x^3 + C, however the requirement that lim [f(x)/x] = 0 implies that C=0.
     
  14. Sep 10, 2011 #13

    NascentOxygen

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    I find this function intriguing. I've been mulling this over since it was first posted, and I'm no nearer to understanding what it is I'm looking at.

    (I understand the OP's question, and a bit of playing around with algebra soon answers his question.) My question is more fundamental: what does this f(x+y) look like? How can I represent it graphically?

    Normally, a function of two variables is written f(x,y). This one takes only one parameter, so I guess that means it is a function of only one variable, e.g., f(w), yet it involves both x and y. Are x and y both independent variables? Is so, why isn't it a function of two variables, then?

    Is this something in 3 dimensions, e.g., such that on the z axis we plot f(x+y)?

    Maybe it's a recursive function, and y = f(x+y) ??

    I'm hoping someone can clue me in on what's going on here.
     
  15. Sep 10, 2011 #14

    D H

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    @Nascent: Think of g(x,y) = sin(x+y). There's nothing confusing about that, is there?

    Or think of the definition of the derivative, the limit of (f(x+h)-f(x))/h as h tends to zero. There's nothing special here about that variable h. The limit of (f(x+y)-f(x))/y as y tends to zero is also f'(x). Once again we have a function of one variable, f(x) but we are evaluating it at the point x+h (or at the point x+y).
     
  16. Sep 10, 2011 #15

    NascentOxygen

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    You haven't specified what y is.

    I'm perfectly happy with y = g(x,w) = sin(x+w)
    where x and w are independent variables.

    If the normal convention applies, i.e., y=g(arg1,arg2) then I remain puzzled, because it's
    y = g(x,y) = sin(x+y)

    So, what precisely is y?
     
  17. Sep 10, 2011 #16

    D H

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    What is x? It isn't specified, either. It is a variable. So is y.
     
  18. Sep 10, 2011 #17

    NascentOxygen

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    On the face of it, I'd expect that h would need to have the same units as x.

    I've never thought of it this way, and if y had the same units as x, this may be true. But if y is f(x), then in general it will not have the same dimensions as x, so a difficulty will be trying to add x + y. (x may be metres, and y metres3, for example, if the function is the volume of a sphere of radius x). Perhaps it isn't necessary to actually perform the addition, it might all come out in the wash. This will then hinge on being able to expand f(a+b)=...
     
  19. Sep 10, 2011 #18

    HallsofIvy

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    y= sin(x+y) is not a definition, it is a "functional equation".
     
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