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Math Help - Prove that f has a limit

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    Prove that f has a limit

    I just posted a similar problem, however this one states:

    define f : (0,1) -> R by f(x)= ((9-x)^(1/2)-3)/x. Prove that f has a limit at 0 and find it.

    This one also states to not use \epsilon and \delta. However, since it says prove it and then find it, how do i do that without an \epsilon, \delta proof?
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tn11631 View Post
    I just posted a similar problem, however this one states:

    define f : (0,1) -> R by f(x)= [tex]sqrt{x-9}[tex]-3/x. Prove that f has a limit at 0 and find it.

    This one also states to not use \epsilon and \delta. However, since it says prove it and then find it, how do i do that without an \epsilon, \delta proof?
    Hint: think derivatives (but I would guess that derivatives may not have been formalized yet)
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    Quote Originally Posted by Drexel28 View Post
    Hint: think derivatives (but I would guess that derivatives may not have been formalized yet)
    sorry I had the problem a little messed up, I've fixed it. However, in class I don't think we've done any like that, or in fact the only ones we did were by using epsilon and delta and now were told not to.
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    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by tn11631 View Post
    sorry I had the problem a little messed up, I've fixed it. However, in class I don't think we've done any like that, or in fact the only ones we did were by using epsilon and delta and now were told not to.
    Then I don't know what prove means
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    Quote Originally Posted by Drexel28 View Post
    Then I don't know what prove means
    lol thats y I am confused. the other one I posted, they suggested to just factor the numerator and I guess plug in 1 normally, but again that problem said prove that f has a limit at 1. But thats kind of just showing it. But this time is says prove the limit exists and find it. so would I just go straight through since i can't use epsilon and delta? or is there another way to prove it
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    Quote Originally Posted by Drexel28 View Post
    Then I don't know what prove means
    ah i think i've got it. . we have (9-x)^(1/2) -3/x and we mult both num and denom by (9-x)^(1/2) + 3 and we get 1/(9-x)^(1/2)+3 and 1 is a constant at each point and the denom at 0 has a limit of 6. therfore there is a theorem that says the limit exits and its 1/6...Guess I just needed a little discussion to get my brain to start lol
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