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Limit and Domain

  1. Apr 4, 2013 #1
    1. The problem statement, all variables and given/known data

    Find the limit (if it exists) and domain of the following function. (I do not need to use epsilon delta proof, just results on different curves to prove this)

    [itex] \stackrel{lim}{_{(x,y)→(−1,0)}}[/itex] [itex] \frac {x(x+1)y^2}{(x+1)^2+y^2}[/itex]

    2. Relevant equations

    3. The attempt at a solution


    Just by looking at function I can see that Domain is [itex]ℝ | x≠0, x≠-1, y≠0 , y^2≠-(x+1)^2|[/itex]

    Having graphed that function in Google I can see that at [itex](-1,0)[/itex] the function is not continuous and if I try and substitute the x and y values into function I end up with [itex]\frac{0}{0}[/itex] and I get similar results if I try [itex]y=mx[/itex]

    I have also tried
    [itex] \stackrel{lim}{_{x→0}}[/itex] [itex] \frac {(x-1)(x-1+1)y^2}{(x-1+1)^2+y^2}[/itex] and [itex]y=mx[/itex]

    Regards
     
    Last edited: Apr 4, 2013
  2. jcsd
  3. Apr 4, 2013 #2

    CompuChip

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    You may want to look closer at the function to find the domain, for example why is x = 0 not allowed, or y² = (x + 1)² ?

    If you substitute y = mx and let x go to -1 you should not get 0/0. Can you show us your calculation in a little more detail?
     
  4. Apr 4, 2013 #3

    SammyS

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    x=0, does not make the denominator zero. Neither does y2 = (x+1)2 make the denominator zero.

    In fact when x = -1, the only case for which the denominator is zero, is when you also have y = 0

    Of course, that's precisely the point at which you need to find the limit.

    By the way, the line given by y = mx does not pass through the point, (-1, 0) . Consider the line y = m(x+1) . Looking at the limit along this line is only helpful if the limit doesn't exist.


    I suggest shifting you curve one unit to the right., and the switching to polar coordinates.
     
  5. Apr 4, 2013 #4
    Thanks for your replies,

    I missed a - sign in my OP, so the denominator goes to zero if [itex]x=-1[/itex] [itex]y=0[/itex] or [itex]y^2=-(x+1)^2[/itex] and numerator goes to zero if [itex]x=0[/itex] or [itex]x=-1[/itex]


    By shifting curve to the right I get the following for the function [itex]\frac{(x-1)xy^2}{x^2+y^2}[/itex] and by letting [itex]y=m(x+1)[/itex] I get [itex] \frac {(x-1)xm^2(x+1)^2}{x^2+m^2(x+1)^2} [/itex] I can't seem to take it any further from here on and subbing 0 for [itex] x[/itex] I get [itex] \frac {0}{m^2}[/itex]

    where have I gone wrong in try to find the limit ( or showing that it does not exist)?

    Regards
     
    Last edited: Apr 4, 2013
  6. Apr 4, 2013 #5

    CompuChip

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    For [itex]m \neq 0[/itex], 0/m² = 0.
    All you need to check is that it is also 0 when y = 0, i.e. where you are approaching along the x-axis.

    I thought you were talking about the function over the reals by the way, in which case [itex]y^2 = -(x + 1)^2[/itex] will never hold (except for when x = -1 and y = 0 as you already noted), because y² and (x + 1)² are always positive.
     
  7. Apr 4, 2013 #6

    SammyS

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    Which sign did you change?

    Now that you've shifted the function you need to do the limit of the shifted function as (x, y) (0, 0), so you should use ##\ y = mx\ .\ ## Unless I've messed something up, the resulting limit does exist. That being the case, the problem may appear to be solved, but it's not solved. Such a method can only show (in some cases) that a limit does not exist.

    Change ##\displaystyle \ \frac{(x-1)xy^2}{x^2+y^2}\ ## to polar coordinates, for which you merely need to consider the limit as r → 0 .
     
  8. Apr 4, 2013 #7
    Thanks for your replies, I am a bit rusty with polar coordinates but using [itex]y=mx[/itex] I get the following with the function.

    [itex] \frac {(x-1)xm^2x^2}{x^2+m^2x^2}[/itex]

    [itex] \frac {(x-1)xm^2x^2}{x^2(m^2+1)} [/itex]

    [itex] \frac {(x-1)xm^2}{m^2+1}[/itex]

    [itex] \frac {m^2x^2-m^2x}{m^2+1}[/itex]


    is this the right method? also having a quick read on polar coordinates I see that [itex]x^2+y^2=r^2[/itex] how can I change [itex] (x-1)xy^2[/itex] to polar form?

    Regards
     
  9. Apr 4, 2013 #8

    Mark44

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    The standard conversion to polar coordinates is:
    x = r cos(θ)
    y = r sin(θ)
    from which we get x2 + y2 = r2

    On the line y = mx, y = m*r*cos(θ).

    BTW, instead of using SIZE to make itex stuff larger, just use tex without the SIZE tags. It's also better, in my opinion, to use the least amount of LaTeX that you can, as large tracts of LaTeX can bog some browsers down, making the pages slow to render. None of what I wrote above uses LaTeX.
     
  10. Apr 4, 2013 #9

    SammyS

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    Have you taken the limit of [itex]\displaystyle \ \frac {m^2x^2-m^2x}{m^2+1}\ [/itex] as x → 0 ?

    While approaching (0, 0) along the line y = mx can be useful to show that a limit doesn't exist, it can't show that a limit (x,y) → (0,0) does exist, although if the limit does exist, this will show you what the limit is.

    When you change to polar coordinates do it with the expression ## \displaystyle \ \frac{(x-1)xy^2}{x^2+y^2}\,,\ ## not with [itex]\displaystyle \ \frac {m^2x^2-m^2x}{m^2+1}\ .[/itex]
     
  11. Apr 4, 2013 #10
    Thanks, so in polar coordinates the function can be written as [itex]\displaystyle \ \frac{(rcosθ-1)rcosθ(rsinθ)^2}{r^2}[/itex]

    and just subbing [itex]\displaystyle r=0[/itex] yields to [itex]\displaystyle \frac {0}{0}[/itex] or do I need to further simplify the function in polar form and then sub in [itex]\displaystyle r=0[/itex]

    Also for the domain I can see as you mentioned before (for the denominator part) [itex]\displaystyle (x+1)^2[/itex] can not be a negative number as it is square and same applies to [itex]\displaystyle y^2[/itex] would the domain be all [itex]\displaystyle ℝ[/itex] , [itex]\displaystyle y≠0[/itex] or [itex]\displaystyle x≠-1,0[/itex]

    Regards
     
  12. Apr 4, 2013 #11

    SammyS

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    How much work have you done with limits?

    What sort of answer is 0/0 for a limit or for anything else?

    Factor whatever power of r that you can from the numerator & cancel whatever possible from the denominator.

    How much work have you done with limits?
     
  13. Apr 4, 2013 #12
    a little bit, but this one is confusing me for some reason (possibly lack of experience with limits)


    ## \frac {0}{0} ## would be undermined, right?

    so this is how I have gone with factoring it:

    ## \frac{(r cos θ -1)r cos θ (r sin θ)^2}{r^2}
    = \frac {(r^2 cos^2 θ - r cos θ)r^2 sin ^2 θ}{r^2}
    = \frac {r((r cos^2 θ - cos θ)r sin^2 θ)}{r^2}
    = \frac {(r cos^2 θ - cos θ)r sin^2 θ}{r}
    = (r cos^2 θ - cos θ) sin^2 θ ##

    and setting ##r=0## i get ## -cos θ sin^2 θ## do I then use trig identity ## sin^2 θ = \frac {1-cos 2θ}{2}## ? also as ##r→0## does ##θ## also tend to ##0##?

    Regards
     
  14. Apr 4, 2013 #13

    SammyS

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    ##\displaystyle \ \frac{(rcosθ-1)rcosθ(rsinθ)^2}{r^2}##

    is the same as

    ##\displaystyle \ \frac{r^3(r\cosθ-1)\cosθ(\sin^2θ)}{r^2}\quad\to\quad r(r\cosθ-1)\cosθ( \sin^2θ)\ .##

    As r → 0, this is 0, for all values of θ . Right ?
     
  15. Apr 4, 2013 #14

    Thanks, so the limit is at [itex](-1,0) [/itex] is 0 regardless of the path taken ( which would be represented by θ) right?

    also did I have the right idea in regards to the domain of the funtion? all ##ℝ## ##|y≠0## , ##x≠-1,0|##
    Regards
     
  16. Apr 4, 2013 #15

    SammyS

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    The original function, ##\displaystyle \ \frac {x(x+1)y^2}{(x+1)^2+y^2}\ ##, is defined at x=0 and y=0, i.e. at (x,y)=(0,0), and is 0 at that point.

    ##\displaystyle \frac{0(0+1)0^2}{(0+1)^2+0^2}=\frac{0}{1}=0\ ##
     
  17. Apr 4, 2013 #16
    Given that (0,0) is defined would I be correct in saying the function in undefined only at (-1,0) hence ##D= R## except ##x=-1, y=0## ?

    Regards
     
  18. Apr 4, 2013 #17

    SammyS

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    \mathbb{R}

    The domain is all of ##\ \mathbb{R}^2\,,\ ## except (-1, 0) .
     
  19. Apr 5, 2013 #18
    Thanks for your help
     
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