# Math Help - Question

1. ## 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!

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3. ## Re: Question

Originally Posted by davidciprut
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?
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
No, it's like saying delta is less than or equal to 1. We still have not said exactly what delta is.

, 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!
We started by 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.

4. ## Re: Question

Originally Posted by davidciprut
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.
We must deal with
That factors as

We have control over the second factor so we use it to bound the first factor. The use of 1 is purely for convince. In fact you could use any positive number.

Originally Posted by davidciprut
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..)
Because it may be the case that