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- February 8th 2014, 02:51 AM #1

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## Question

I have uploaded a question, that I have encountered million times, however I still have problem understanding a point in it.

It says ''If we insist |x-2|<delta then 1<x<3 and so |x+2|<5''

The assumption that |x-2|<1 means we choose delta=1. Right?

So therefore at the and we get that

|x^2-4|=|x+2||x-2|<5|x-2|<5delta=epsilon and we get from here that delta=epsilon/5 and therefore they said delta=min{1,epsilon/5}

Now the thing that I have trouble understanding is, when we assume that |x+2|<5, we already chose delta to be one, didn't we? So how come we can say that

5|x-2|<5delta , delta is one in this case no?Why do we still write delta? What am I not getting here?

It's like saying lets assume delta=1, then we get |x+2|<5, and then when we go back to the inequality and we say the inequality smaller then 5delta, as if delta is equal to 1 and another number at the same time. I hope I was clear enough to explain the point that I didn't get.

And the other thing that I didn't understand why do we chose the minimum between 1 and epsilon/5? (Maybe I will understand this part if I get the first part..)

Thank you!

- February 8th 2014, 09:19 AM #2

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- February 8th 2014, 10:04 AM #3

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## Re: Question

No. Of course, |x- 2|< delta is the same as -delta< x- 2< delta, 2- delta< x< 2+ delta so that if delta= 1 those will be true. But then it will be true for any

**smaller**value of delta as well. This is saying delta is**less than or equal to**1.

So how come we can say that

5|x-2|<5delta , delta is one in this case no?Why do we still write delta? What am I not getting here?

It's like saying lets assume delta=1, then we get |x+2|<5

, and then when we go back to the inequality and we say the inequality smaller then 5delta, as if delta is equal to 1 and another number at the same time. I hope I was clear enough to explain the point that I didn't get.

And the other thing that I didn't understand why do we chose the minimum between 1 and epsilon/5? (Maybe I will understand this part if I get the first part..)

Thank you!**asserting**that delta was less than or equal to 1. We then derived that, to satisfy the inequalities needed (for continuity?) we must also have x< epsilon/5. Since the second inequality follow from the first, they must**both**be true. Because of the transitive property of inequality (if a< b and b< c then a < c) if x is less than**the smaller**of the two, it is smaller than both.

- February 8th 2014, 10:33 AM #4
## Re: Question