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    Prove that root(6-x) tends to 2 as x tends to 2
    [Hint: Multiply the top and bottom by root(6-x) + 2

    Definition: (For all epsilon>0) (There exists delta>0) (For all x E R \ {x</6})
    if 0<|x-2|< delta implies |root(6-x) - 2| < epsilon

    Taking |root(6-x) - 2| we times both sides by root(6-x) + 2 giving| (2-x)/(root6-x)+2| = |x-2| / | (root 6-x) + 2|

    So |x-2| / | (root 6-x) + 2| < |x-2| so let delta = epsilon.


    Is this all accurate? Obviously from the delta = epsilon its obvious...


    I hope you can decipher my awful writing on here!
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      (Original post by Cggh90)
      x
      I have only had a very brief look, but what you've written looks mostly okay. The following line is a bit confusing as you've clearly done some working that isn't shown, but I presume you've written it out in my detail on paper!:

      "Taking |root(6-x) - 2| we times both sides by root(6-x) + 2 giving| (2-x)/(root6-x)+2| = |x-2| / | (root 6-x) + 2|"

      Also, there's a typo in this line:

      "So |x-2| / | (root 6-x) + 2| < |x-2| so let delta = epsilon." It should obviously have &lt;\epsilon rather than &lt;|x-2|.
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      (Original post by Kolya)
      I have only had a very brief look, but what you've written looks mostly okay. The following line is a bit confusing as you've clearly done some working that isn't shown, but I presume you've written it out in my detail on paper!:

      "Taking |root(6-x) - 2| we times both sides by root(6-x) + 2 giving| (2-x)/(root6-x)+2| = |x-2| / | (root 6-x) + 2|"

      Also, there's a typo in this line:

      "So |x-2| / | (root 6-x) + 2| < |x-2| so let delta = epsilon." It should obviously have &lt;\epsilon rather than &lt;|x-2|.
      Yeh, I meant epsilon.

      Ther trouble I had was preventing root(6-x) from being negative.

      Was it ok to say in the definition xER \ {x</6} ???
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        (Original post by Cggh90)
        Yeh, I meant epsilon.

        Ther trouble I had was preventing root(6-x) from being negative.

        Was it ok to say in the definition xER \ {x&lt;/6} ???
        Yeah, maybe just say at the start of your answer that \sqrt{6-x} is not defined on \mathbb{R} when x>6, hence you will apply the definition for x<=6.
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        (Original post by Kolya)
        Yeah, maybe just say at the start of your answer that \sqrt{6-x} is not defined on \mathbb{R} when x>6, hence you will apply the definition for x<=6.
        Thanks alot

        Are you able to help me with one more?


        Prove (from the definition) that if f(x) tends to L as x tends to a then 2f(x) tends to 2L as x tends to a

        Now we can suppose (For all epsilon>0) (There exists delta >0) (For all x E X)
        0<|x-a|<delta implies |f(x) - L| < epsilon.

        Now we need to prove that 2f(x) tends to 2L as x tends to a.

        I'm struggling a bit with how to do this..
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          (Original post by Cggh90)
          I'm struggling a bit with how to do this..
          A typical trick, which I'm sure you'll have seen used in different examples in your book/lectures, is to say just change the notation a little and write something like:

          "Now we can suppose (For all \epsilon _1 &gt;0) (There exists delta >0) (For all x E X)
          0<|x-a|<delta implies |f(x) - L| &lt; \epsilon _1."

          Maybe go back and see how those kinds of proofs were done, and think about how you can apply similar ideas to this question?
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          (Original post by Kolya)
          A typical trick, which I'm sure you'll have seen used in different examples in your book/lectures, is to say just change the notation a little and write something like:

          "Now we can suppose (For all \epsilon _1 &gt;0) (There exists delta >0) (For all x E X)
          0<|x-a|<delta implies |f(x) - L| &lt; \epsilon _1."

          Maybe go back and see how those kinds of proofs were done, and think about how you can apply similar ideas to this question?
          I've still not managed this

          You've just changed epsilon to epsilon 1?

          Can you give me more of a hint!
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          (Original post by Cggh90)
          I've still not managed this

          You've just changed epsilon to epsilon 1?

          Can you give me more of a hint!
          Your work almost right but you should to take
          some restriction to x and delta
          As it was written x \le 6 \rightarrow \delta \le 4
          as x tend to 2
          And the coclusion at the end
          \delta \le min {4,\epsilon}
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            (Original post by Cggh90)
            I've still not managed this

            You've just changed epsilon to epsilon 1?

            Can you give me more of a hint!
            The next stage is to consider |2f(x) - 2L|. You know that there exists a delta such that |f(x) - L| &lt; \epsilon _1, so you will be able to find a bound for |2f(x) - 2L| as well (using the same delta).
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            (Original post by Kolya)
            The next stage is to consider |2f(x) - 2L|. You know that there exists a delta such that |f(x) - L| &lt; \epsilon _1, so you will be able to find a bound for |2f(x) - 2L| as well (using the same delta).
            I did this:

            Suppose (For all epsilon > 0) (There exists delta > 0) (For all x E X) 0< |x-1| < delta implies |f(x) - L| < epsilon

            As epsilon > 0, epsilon/2 > 0 so

            (For all epsilon/2 > 0) (There exists delta > 0) (For all x E X) 0< |x-1| < delta implies |f(x) - L| < epsilon/2


            |f(x)-L| < epsilon / 2 can be written |2f(x) - 2L| < epsilon

            So (For all epsilon > 0) (There exists delta > 0) (For all x E X) 0< |x-1| < delta implies |2f(x) - 2L| < epsilon



            I kind of know this isn't sufficient. Could you tweak it for me!
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              (Original post by Cggh90)
              I did this:

              Suppose (For all epsilon &gt; 0) (There exists delta &gt; 0) (For all x E X) 0&lt; |x-1| &lt; delta implies |f(x) - L| &lt; epsilon

              As epsilon &gt; 0, epsilon/2 &gt; 0 so

              (For all epsilon/2 &gt; 0) (There exists delta &gt; 0) (For all x E X) 0&lt; |x-1| &lt; delta implies |f(x) - L| &lt; epsilon/2


              |f(x)-L| &lt; epsilon / 2 can be written |2f(x) - 2L| &lt; epsilon

              So (For all epsilon &gt; 0) (There exists delta &gt; 0) (For all x E X) 0&lt; |x-1| &lt; delta implies |2f(x) - 2L| &lt; epsilon



              I kind of know this isn't sufficient. Could you tweak it for me!
              Yeah, you're pretty much there. Although: firstly, i think you have a typo. it should be |x-a| not |x-1|. Secondly, perhaps the neatest and clearest way to write it would be:

              Suppose (For all epsilon_1 > 0) (There exists delta > 0) (For all x E X) 0< |x-1| < delta implies |f(x) - L| < epsilon_1

              Whenever epsilon_1 > 0, we have 2(epsilon_1) > 0. It follows that:

              (For all 2(epsilon_1) > 0) (There exists delta > 0) (For all x E X) 0< |x-1| < delta implies |f(x) - L| < epsilon_1

              Define epsilon = 2(epsilon_1). Then substituting in tells us:

              (For all epsilon > 0) (There exists delta > 0) (For all x E X) 0< |x-1| < delta implies |f(x) - L| < (epsilon)/2

              i.e.

              (For all epsilon > 0) (There exists delta > 0) (For all x E X) 0< |x-1| < delta implies |2f(x) - 2L| < epsilon


              This makes all the reasoning explicit.

              imho, your working is a little unclear. the problem is that it's not 100% obvious, at least not to me, that (epsilon > 0 => epsilon/2 > 0) implies that: (For all epsilon/2 > 0) (There exists delta > 0) (For all x E X) 0< |x-1| < delta implies |f(x) - L| < epsilon/2

              You seem to be mixing a substitution and an implication.

              If instead you use a substitution in one step, and use the implication (epsilon > 0 => 2epsilon > 0) in another step, then it becomes much clearer - at least to me! Either way, perhaps writing some more explanation of what you're doing, rather than just using the word "so", might be helpful.

              (nb. i'm really not great at this stuff, so it's possible i'm flagging up stuff that is completely fine...)
             
             
             
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