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Epsilon and delta definition of limit

  1. Apr 24, 2006 #1
    Can anyone help me on this question,Using [tex]\epsilon[/tex]-[tex]\delta[/tex] definition of a limit to show that

    [tex]\lim[/tex] [tex]\frac {2x^3-y^3}{x^2+y^2}[/tex] = 0
  2. jcsd
  3. Apr 25, 2006 #2


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    Like with most of these, I recommend changing to polar coordinates.
    [itex]x= r cos(\theta)[/itex] and [itex]y= r sin(\theta)[/itex]
    [tex]\frac{2x^3-y^3}{x^2+y^2}= \frac{r^3(cos^3(\theta)- sin^3(\theta))}{r^2}= r(cos^3(\theta)- sin^3(\theta))[/tex]. Now the distance from (0,0) is entirely encapsulated in r.
  4. Apr 25, 2006 #3


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    or just factor out cubrt(2)x-y from the top, which certainly goes to zero as x and y do.

    then it suffices to show the other factor remains bounded. but it obviously does, since the top has three terms, none as much as twice as large as the bottom, i.e. the other factor is bounded by 6.
  5. May 10, 2006 #4
    anybody can explain what is the usage of epsilon delta definition of limit?what will happen without them?
  6. May 10, 2006 #5

    matt grime

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    What is your definition of a limit without using epsilons and deltas?

    Their usage is to formally define limits. If you don't use them you have to give another way of defining limits. What is it? Whatever you come up with it must be equivalent to the epsilon and delta definition (or it will be convergence in some other sense: it is possible to define limits for any topology (which we won't define here), but in the reals we are using the metric topology, that is we are using balls of radii epsilon and delta to define the open sets. Thus whilst it is prefectly possible to not mention espilon or delta in a definition of limit, you are merely glossing over the fact that they are there. For instance, we may say a_n tends to a if every open set containing a contains all but finitely many of the a_n, for sequences, or we can say that f is continuous if for every open set in the range of f the inverse image is open. Both of these avoid mentioning epsilon and delta, but the open sets are defined in terms of open balls or intervals, which is just the epsilon delta thing in disguise. Suppose we wanted to verify that f(x)=x^2 from R to R is continuous, then pick some open interval in the image (a,b), and assume a>0, then the inverse image is (-sqrt(b),-sqrt(a)) union (sqrt(a),sqrt(b), if a<=0 and b>0 then the inverse image is (-sqrt(b),sqrt(b)), if b<0 then the inverse image is the empty set, thus x^2 is continuous. Easier, I think we all agree. Of course if f(x) is something that is not so easily 'inverted' then you're not necessarily going to be any better off than using epsilons and deltas, for instance, try doing f(x)=x^6-x^3+x^2+1)
    Last edited: May 10, 2006
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