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Math Help - limit of arcsin/arctan in two variables

  1. #1
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    limit of arcsin/arctan in two variables

    what is the steps for calculating lim (x,y)->(1,3) [arcsin(3xy-9)/arctan(xy-3)]
    i know how to do it in 1 dimension..if let z=xy. can i actually do this?
    my solution is 3

    actually how to use [tex] code in this forum?
    i am new here

    thanks!
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  2. #2
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    Quote Originally Posted by pokemon1111 View Post
    what is the steps for calculating lim (x,y)->(1,3) [arcsin(3xy-9)/arctan(xy-3)]
    i know how to do it in 1 dimension..if let z=xy. can i actually do this?
    my solution is 3

    actually how to use [tex] code in this forum?
    i am new here

    thanks!
    Yes. Yes. Click on the relavant link in my signature.
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  3. #3
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    Quote Originally Posted by mr fantastic View Post
    Yes. Yes. Click on the relavant link in my signature.
    New users (anyone with less than 10 posts) can't view links in signatures. (Why? I have no clue.)

    The link for the latex tutorial is here: http://www.mathhelpforum.com/math-he...-tutorial.html
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  4. #4
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    thanks all
    why can i do so? i mean from 2 var->1 variable
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  5. #5
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    Since x and y only appear as "xy", you can make that substitution. No matter how (x, y) approaches (1, 3), z= xy approaches 3.

    If it had been something like "[arcsin(3xy-9)/arctan(y-3)]" you would not have been able to do that.
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  6. #6
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    Quote Originally Posted by HallsofIvy View Post
    Since x and y only appear as "xy", you can make that substitution. No matter how (x, y) approaches (1, 3), z= xy approaches 3.

    If it had been something like "[arcsin(3xy-9)/arctan(y-3)]" you would not have been able to do that.
    what should i write to express this?
    just write let z=xy, when (x,y)->(1,3), z->3
    then continue like one var?

    if something like this, how can i calculate?
    thanks~!
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  7. #7
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    You would just say something like

    Let z=xy. So if (x,y) \to (1,3), then z \to 3.

    Therefore:

    \lim_{(x,y)\to(1,3)} \frac{\arcsin(3xy-9)}{\arctan(xy-3)} = \lim_{z \to 3} \frac{\arcsin(3z-9)}{\arctan(z-3)} = \cdots

    You could continue solving the limit using standard one variable techniques.
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  8. #8
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    thank you very much =)
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