# Applying the epsilon-delta definition

## Homework Statement

Consider the function g(x) = 1/x.
b. Find a number δ′ > 0 different from the number δ you found in the previous question such that all the values x ∈ (2 − δ′, 2 + δ′) satisfy that g(x) is within 0.1 of 1/2.

Note, I found I value δ>0 in an earlier question in which the values x∈(2−δ,2+δ) satisfy that g(x) is within 0.1 of 1/2. That value of δ=1/10.

This question b. practically asks the same thing, but for a different value δ'.

## The Attempt at a Solution

Now, I'm considering since we know δ=1/10, then 0<δ'<δ=1/10. Any value of δ' in this interval will satisfy the conditions in question. But how would I show this, more formally.

jbunniii
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## Homework Statement

Consider the function g(x) = 1/x.
b. Find a number δ′ > 0 different from the number δ you found in the previous question such that all the values x ∈ (2 − δ′, 2 + δ′) satisfy that g(x) is within 0.1 of 1/2.

Note, I found I value δ>0 in an earlier question in which the values x∈(2−δ,2+δ) satisfy that g(x) is within 0.1 of 1/2. That value of δ=1/10.

This question b. practically asks the same thing, but for a different value δ'.

## The Attempt at a Solution

Now, I'm considering since we know δ=1/10, then 0<δ'<δ=1/10. Any value of δ' in this interval will satisfy the conditions in question. But how would I show this, more formally.
See if you can find exactly what range of values ##x## satisfy the requirement. You need
$$\left|\frac{1}{x} - \frac{1}{2}\right| < \frac{1}{10}$$
This is equivalent to
$$-\frac{1}{10} < \frac{1}{x} - \frac{1}{2}< \frac{1}{10}$$
Now do some algebra to try to turn this into an equivalent inequality of the form
$$??? < x < ???$$

And so I obtain 5/3 <x<5/2.

But how does this demonstrate that 0<δ'<1/10?

Opps I just realized I made an error with a previous calculation, so neglect what I said earlier.

Last edited:
jbunniii
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Which number (5/3 or 5/2) is closer to 1/2?

5/3 is closer.
I was thinking maybe I should just solve for δ' the same way I solved for δ (by using the epsilon delta definition, where ε = 1/10. And solving for δ' the same way you'd prove a limit).
Would that be simpler?

Office_Shredder
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Now, I'm considering since we know δ=1/10, then 0<δ'<δ=1/10. Any value of δ' in this interval will satisfy the conditions in question. But how would I show this, more formally.

I like this approach a lot. Let's pick some δ', let's say δ' = 1/20. Then you want to prove that if 2-1/20 < x < 2+1/20, then |1/x-1/2| < .1. You should be able to do this in a single line knowing what you know.

So, letting δ'=1/20, and if |x-2|<1/20 then |1/x-1/2|<1/10. But from this what do we want to prove? There is no value δ to solve for.

Last edited:
jbunniii
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5/3 is closer.
I was thinking maybe I should just solve for δ' the same way I solved for δ (by using the epsilon delta definition, where ε = 1/10. And solving for δ' the same way you'd prove a limit).
Would that be simpler?
Oops, I meant: which number is closer to 2, not which one is closer to 1/2. OK, so 5/3 is closer to 2. The distance between 5/3 and 2 is 1/3. So any ##x## which is closer than 1/3 to the point ##x=2## will work. What does this imply about the range of possible values for ##\delta'##?

(And yes, your proposed approach will work too. My approach will allow you to find all possible values of ##\delta'## that solve the problem, but that may be overkill.)

Office_Shredder
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So, letting δ'=1/20, and if |x-2|<1/20 then |1/x-1/2|<1/10. But from this what do we want to prove? There is no value δ to solve for.

No, δ'=1/20 is the value that you solved for in your very first post. You said that any value δ' < 1/10 should work, so I said well let's take δ'=1/20 and see that it does work.

jbunniii
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So, letting δ'=1/20, and if |x-2|<1/20 then |1/x-1/2|<1/10. But from this what do we want to prove? There is no value δ to solve for.
Well, 1/20 < 1/10, so if ##|x-2| < 1/20## then ##|x-2| < 1/10##. But 1/10 is the ##\delta## from your previous problem. Therefore...

Well, 1/20 < 1/10, so if ##|x-2| < 1/20## then ##|x-2| < 1/10##. But 1/10 is the ##\delta## from your previous problem. Therefore...

I should probably note, that I made an error in calculating δ in the previous question. Its value should be δ=1/5. Although, it wouldn't make a huge difference.

So putting everything together:

We know from the previous question that δ=1/10 works. Then δ' will also work if
0<δ'<δ=1/10.
To prove this suppose δ'=1/20. Then, 1/20<1/10. So it follows that if |x-2|<1/10 then |x-2|<1/20. But we know δ works, as verified in the previous question. Therefore, δ' will also work.

Mark44
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I should probably note, that I made an error in calculating δ in the previous question. Its value should be δ=1/5. Although, it wouldn't make a huge difference.

So putting everything together:

We know from the previous question that δ=1/10 works. Then δ' will also work if
0<δ'<δ=1/10.
To prove this suppose δ'=1/20. Then, 1/20<1/10. So it follows that if |x-2|<1/10 then |x-2|<1/20.
No, this doesn't follow. If |x - 2| < 1/10, then x ##\in## (1.9, 2.1). There are plenty of numbers in this interval that aren't within .05 (= 1/20) of 2. A couple of them are x = 1.91 and 2.09.
But we know δ works, as verified in the previous question. Therefore, δ' will also work.

No, this doesn't follow. If |x - 2| < 1/10, then x ##\in## (1.9, 2.1). There are plenty of numbers in this interval that aren't within .05 (= 1/20) of 2. A couple of them are x = 1.91 and 2.09.

Opps it should be if |x-2|<1/20, then |x-2|<1/10.