Page 1 of 2 12 LastLast
Results 1 to 15 of 26

Math Help - Squeeze Rule? R2 limit

  1. #1
    Member
    Joined
    Aug 2007
    Posts
    239

    Squeeze Rule? R2 limit

    Can you use the squeeze rule for the following:  \lim_{(x,y) \to (0,0)} \frac{x^{2}+y^{2}}{\sqrt{x^{2}+y^{2}+1}-1} rather than multiplying by the conjugate?

    I did the following:  0 \leq \frac{x^{2}+y^{2}}{\sqrt{x^{2}+y^{2}+1}-1} \leq  x^{2} + y^{2} . So wouldn't the limit be  0 ?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,923
    Thanks
    1762
    Awards
    1
    Quote Originally Posted by shilz222 View Post
     0 \leq \frac{x^{2}+y^{2}}{\sqrt{x^{2}+y^{2}+1}-1} \leq  x^{2} + y^{2} .
    Try that inequality for x=0.1 & y=0.2.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Aug 2007
    Posts
    239
    What a dumb mistake. Aargh.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,923
    Thanks
    1762
    Awards
    1
    Can you use polar substitution?
    r^2  = x^2  + y^2 ,\;x = r\cos (t),\;y = r\sin (t).

    You will get a function of r alone and then let r \to 0
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Member
    Joined
    Aug 2007
    Posts
    239
    yes it would be  \frac{r^{2}}{\sqrt{r^2+1}-1} so now you can use L'Hopitals rule.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,923
    Thanks
    1762
    Awards
    1
    Very good. Is the limit 2?
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Member
    Joined
    Aug 2007
    Posts
    239
    yes. the limit is 2. I made a mistake on that one on a test.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Just rationalize!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Aug 2007
    Posts
    239
    Yes I missed that. I knew I should have done that. But alas, I was under a time constraint.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Member
    Joined
    Aug 2007
    Posts
    239
    Can you say the limit is 0 if we restrict  x \in \mathbb{Z} ? Or is this not allowed?
    Follow Math Help Forum on Facebook and Google+

  11. #11
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by shilz222 View Post
    Can you say the limit is 0 if we restrict  x \in \mathbb{Z} ? Or is this not allowed?
    it's not allowed. i think the question was meant for real numbers, otherwise, we could not make quantities arbitrarily close to each other and the limit would make no sense then
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Aug 2007
    Posts
    239
    But we have Real Analysis and we have Complex Analysis. So why cant we consider the limit in  \mathbb{Z}^{n} ? (i.e. Integer Analysis)? Couldn't we also consider this same limit in  \mathbb{C}^{n} ? It wouldnt make sense in  \mathbb{R}^{n} but would it make sense in  \mathbb{Z}^{n} ?
    Follow Math Help Forum on Facebook and Google+

  13. #13
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by shilz222 View Post
    But we have Real Analysis and we have complex analysis. So why cant we consider the limit in  \mathbb{Z}^{n} ? (i.e. Integer Analysis)? Couldn't we also consider this same limit in  \mathbb{C}^{n} ?
    \mathbb{R} and \mathbb{C} are complete sets (the completeness axiom applies). this is a property that makes it possible to evaluate limits, because we need to make things as close as possible. this is not possible with the integers, since there are gaps 1 unit in length between integers. finding limits under those conditions make no sense really, since we can't make things arbitrarily close (recall the epsilon-delta definition of the limit of a function)
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Member
    Joined
    Aug 2007
    Posts
    239
    But the definition of the limit (epsilon-delta) is based on  \mathbb {R}^{n} ? So we cant really use a definition of a limit that is defined in  \mathbb{R}^{n} in  \mathbb{Z}^{n} right? So basically when you take a limit in  \mathbb{Z}^{n} the closest you can be is 1 unit away. So cant you restrict the definition of a limit? You are still taking a limit right?

    Actually I think you can.  \mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset  \mathbb{R} by the Dedekind cuts. So the definition that applies in  \mathbb{R}^{n} will also apply to  \mathbb{Z}^{n} but not vice-versa.
    Follow Math Help Forum on Facebook and Google+

  15. #15
    is up to his old tricks again! Jhevon's Avatar
    Joined
    Feb 2007
    From
    New York, USA
    Posts
    11,663
    Thanks
    3
    Quote Originally Posted by shilz222 View Post
    But the definition of the limit (epsilon-delta) is based on  \mathbb {R}^{n} ? So we cant really use a definition of a limit that is defined in  \mathbb{R}^{n} in  \mathbb{Z}^{n} right? So basically when you take a limit in  \mathbb{Z}^{n} the closest you can be is 1 unit away. So cant you restrict the definition of a limit? You are still taking a limit right?

    Actually I think you can.  \mathbb{N} \subset \mathbb{Z} \subset \mathbb{Q} \subset  \mathbb{R} by the Dedekind cuts. So the definition that applies in  \mathbb{R}^{n} will also apply to  \mathbb{Z}^{n} but not vice-versa.
    we can take limits in \mathbb{C} as well. even if you could come up with a construction for limits in the integers, it would be essentially meaningless. since it would give rise to things like continuity at a point where there is actually a gap. it would simply make no sense, and won't be very interesting. in any case, if you considered only integers here, the answer would still be 2, so why even bother with it?
    Follow Math Help Forum on Facebook and Google+

Page 1 of 2 12 LastLast

Similar Math Help Forum Discussions

  1. Replies: 6
    Last Post: March 4th 2011, 09:27 AM
  2. Evaluate the limit using the squeeze theorem?
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 17th 2009, 01:03 PM
  3. Limit without L'Hospital's Rule.
    Posted in the Calculus Forum
    Replies: 4
    Last Post: January 11th 2009, 02:02 PM
  4. Using the Squeeze Theorem to find a limit.
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 28th 2008, 12:14 AM
  5. limit with l'hospitals rule
    Posted in the Calculus Forum
    Replies: 6
    Last Post: November 18th 2007, 06:07 PM

Search Tags


/mathhelpforum @mathhelpforum