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Math Help - Limits and continuity in multivariable calculus

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    Limits and continuity in multivariable calculus

    Hey,
    I'm having quite some trouble FULLY understanding this topic, and in particular this definition is difficult to get my head around:

    What is the direct approach to take when trying to evaluate whether the limit exists or not?

    Take this example from the textbook:
    http://i.imgur.com/DQ16c6j.jpg

    I don't understand why they've evaluated the limits by first approaching from (0,0) along the x-axis and then the y-axis if they were only going to dismiss this idea by saying "Although we have obtained identical limits along the axes, that does not show that the given limit is 0."
    Why not just approach (0,0) from the x-axis and then (0,0) from y=x straight off the bat to eliminate the unecessary step of approaching (0,0) from the y-axis?
    I feel like I am missing some fundamental idea here, what is it exactly?

    Furthermore, what does it mean "Although we have obtained identical limits along the axes, that does not show that the given limit is 0". If the identical limits along the axes do not show this, what do they tell us then?
    Last edited by 99.95; April 2nd 2013 at 11:17 PM.
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    Re: Limits and continuity in multivariable calculus

    After re-reading the materials a few times, i think I have a better understanding. May someone check my approach to this question, it is different from the book's approach, however I believe more 'correct' ?

    Find the limit, if it exists, or show that the limit does not exist.
    \lim_{(x,y)\to (0,0)}\frac{x^4-4y^2}{x^2+2y^2}

    I let (x,y) --> (0,0) along any non vertical line through the origin, so y = mx, where m is the slope.

    as a result, f(x,y) = f(x,mx) = \frac{x^4 -4(xm)^2}{x^2+2(xm)^2} = \frac{x^2-4m^2}{1+2m^2}

    I wasn't quite sure what this end result means, but my taking is that as 'm' varies, or as we pick different slopes, we get different limits, and hence the limit does not exist?

    The books method was
    approaching (0,0) first from the x-axes and then from the y-axes. As it turns out, the limits were different, but i think my method gives a more generalised answer.
    Am I correct in my method, and reasoning about how as 'm' varies, or as we pick different slopes, we get different limits, and hence the limit does not exist?
    Last edited by 99.95; April 3rd 2013 at 01:16 AM.
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    Re: Limits and continuity in multivariable calculus

    Yes, what you've done using y = mx is approach (0,0) from an infinite amount of directions... not just 2 directions as it's done in 2d. In 3d 2 directions of approach are not enough to determine limits, that's why they showed you 2 directions gave the same answer but a 3rd direction contradicted that result in your first example. Sometimes you can get lucky and pick 2 directions that give different limits, that was their point in your second example.

    In 3d you must approach the point (x,y) from all (infinite amount of) possible directions including y = x^2 etc. in order to determine the limit. Looking at more examples will help you understand better and you'll see how they prove the limit rigorously once the evidence starts to mount that the limit in fact does exist.

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    Re: Limits and continuity in multivariable calculus

    Thanks. Okay, so getting used to limits a bit more now, I have yet another question.

    Find the limit if it exists
    \lim_{(x,y)\to(0,0)}\frac{xy}{\sqrt{(x^2+y^2)}}

    from observation you can concur that (x,y) --> (0,0) from the x axes or y axes will yield 0 and even from y=x^2 or y=x. So you begin to suspect the limit exists.

    I am finding it difficult to use the definition of a limit or squeeze theorem to prove it exists. Could someone please explain step by step how to solve this part?
    Last edited by 99.95; April 3rd 2013 at 02:55 AM.
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    Re: Limits and continuity in multivariable calculus

    Here i will withdraw since i don't have a text book in front of me and i fear any attempt i make using memory will be inaccurate. I hope others will help you.

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