Prove that root(6x) tends to 2 as x tends to 2
[Hint: Multiply the top and bottom by root(6x) + 2
Definition: (For all epsilon>0) (There exists delta>0) (For all x E R \ {x</6})
if 0<x2< delta implies root(6x)  2 < epsilon
Taking root(6x)  2 we times both sides by root(6x) + 2 giving (2x)/(root6x)+2 = x2 /  (root 6x) + 2
So x2 /  (root 6x) + 2 < x2 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!

Cggh90
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 28112010 22:18

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 28112010 23:06
(Original post by Cggh90)
x
"Taking root(6x)  2 we times both sides by root(6x) + 2 giving (2x)/(root6x)+2 = x2 /  (root 6x) + 2"
Also, there's a typo in this line:
"So x2 /  (root 6x) + 2 < x2 so let delta = epsilon." It should obviously have rather than .Last edited by Kolya; 28112010 at 23:08. 
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 28112010 23:18
(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(6x)  2 we times both sides by root(6x) + 2 giving (2x)/(root6x)+2 = x2 /  (root 6x) + 2"
Also, there's a typo in this line:
"So x2 /  (root 6x) + 2 < x2 so let delta = epsilon." It should obviously have rather than .
Ther trouble I had was preventing root(6x) from being negative.
Was it ok to say in the definition xER \ {x</6} ??? 
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 28112010 23:38
(Original post by Cggh90)
Yeh, I meant epsilon.
Ther trouble I had was preventing root(6x) from being negative.
Was it ok to say in the definition xER \ {x</6} ??? 
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 28112010 23:55
(Original post by Kolya)
Yeah, maybe just say at the start of your answer that is not defined on when x>6, hence you will apply the definition for x<=6.
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<xa<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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 29112010 00:18
(Original post by Cggh90)
I'm struggling a bit with how to do this..
"Now we can suppose (For all ) (There exists delta >0) (For all x E X)
0<xa<delta implies ."
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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 29112010 18:38
(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 ) (There exists delta >0) (For all x E X)
0<xa<delta implies ."
Maybe go back and see how those kinds of proofs were done, and think about how you can apply similar ideas to this question?
You've just changed epsilon to epsilon 1?
Can you give me more of a hint! 
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 29112010 22:41
(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!
some restriction to x and delta
As it was written
as x tend to 2
And the coclusion at the end
{}Last edited by ztibor; 29112010 at 22:45. 
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 30112010 00:28
(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! 
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 30112010 00:35
(Original post by Kolya)
The next stage is to consider . You know that there exists a delta such that , so you will be able to find a bound for as well (using the same delta).
Suppose (For all epsilon > 0) (There exists delta > 0) (For all x E X) 0< x1 < 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< x1 < 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< x1 < delta implies 2f(x)  2L < epsilon
I kind of know this isn't sufficient. Could you tweak it for me! 
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 30112010 01:10
(Original post by Cggh90)
I did this:
Suppose (For all epsilon > 0) (There exists delta > 0) (For all x E X) 0< x1 < 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< x1 < 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< x1 < delta implies 2f(x)  2L < epsilon
I kind of know this isn't sufficient. Could you tweak it for me!
Suppose (For all epsilon_1 > 0) (There exists delta > 0) (For all x E X) 0< x1 < 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< x1 < 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< x1 < delta implies f(x)  L < (epsilon)/2
i.e.
(For all epsilon > 0) (There exists delta > 0) (For all x E X) 0< x1 < 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< x1 < 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...)Last edited by Kolya; 30112010 at 01:13.
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