Applying the epsilon-delta definition

[NATURE.M]

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'.

Homework Equations

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]

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'.

Homework Equations

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
$$? < x < ?$$

 

[NATURE.M]

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.

 

[jbunniii]

Which number (5/3 or 5/2) is closer to 1/2?

 

[NATURE.M]

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]

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.

 

[NATURE.M]

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.

 

[jbunniii]

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]

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]

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...

 

[NATURE.M]

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]

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.

 

[NATURE.M]

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.