Prove the Limit Statement: Is There a Way to Find δ Given ε and L?

[Saladsamurai]

Problem

I've just kind of been 'going through the motions,' but I feel like the Precise Definition of a Limit is about to set in.

I think some questions about this example should help.

Prove the Limit Statement:

\lim_{x\rightarrow4}(9-x)=5

Attempt

So by asserting that the limit is indeed '5' we are implying that there exists some \delta such that for all 'x'

0<|x-x_o|<\delta\Rightarrow|f(x)-5|<\epsilon

So:

\-\epsilon<(9-x)-5<\epsilon

-\epsilon-4<-x<\epsilon-4

4-\epsilon<x<\epsilon+4

\therefore

\, -\epsilon<x-4<\epsilon

|x-4|<\epsilon=\delta

Now I am a little confused. Have I actually done anything?

Have I shown that so long as I stay within \delta=\epsilon \text{ of }x_o I can get within a distance of \epsilon of 'L.'

Because that's what i am under the impression I have done. But I am not confident about it.

Thanks

 

[VeeEight]

How about saying that 9 - x is a polynomial function and thus continuous.

 

[Saladsamurai]

How about saying that 9 - x is a polynomial function and thus continuous.

Though I can appreciate the simplicity/brevity, that is not the purpose of the exercise.

it is to use the def'n of a limit.

 

[LCKurtz]

Keep in mind that you want to show that given any \epsilon > 0 you can find a \delta that works.

You've done the hard part. Your argument is what might be called an exploratory argument. You started out to see how to get |(9x-5)-4 | < \epsilon and wound up with | x-4| < \epsilon.

Now to prove the limit statement you wish to prove that given \epsilon > 0 you can find \delta > 0 such that if 0 < |x-4| < \delta then
|(9x-5)-4 | < \epsilon

So start your final writeup like this:

Suppose \epsilon > 0. Let \delta = \epsilon. (You know this is going to work from your exploratory argument, but you don't need to include that here.) Now you proceed: If 0 < |x-4| < \delta then...now write your exploratory argument in reverse and end up with |(9x-5)-4 | < \epsilon. That shows your \delta works and you are done.

 

[HallsofIvy]

`

Problem

I've just kind of been 'going through the motions,' but I feel like the Precise Definition of a Limit is about to set in.

I think some questions about this example should help.

Prove the Limit Statement:

\lim_{x\rightarrow4}(9-x)=5

Attempt

So by asserting that the limit is indeed '5' we are implying that there exists some \delta such that for all 'x'

0<|x-x_o|<\delta\Rightarrow|f(x)-5|<\epsilon

So:

\-\epsilon<(9-x)-5<\epsilon

-\epsilon-4<-x<\epsilon-4

4-\epsilon<x<\epsilon+4

\therefore

\, -\epsilon<x-4<\epsilon

|x-4|<\epsilon=\delta

Now I am a little confused. Have I actually done anything?

Have I shown that so long as I stay within \delta=\epsilon \text{ of }x_o I can get within a distance of \epsilon of 'L.'

Because that's what i am under the impression I have done. But I am not confident about it.

Thanks

That is what is sometimes referred to as "synthetic proof". You have gone from what you want to prove, |f(x)- L|< \epsilon, to your hypothesis, |x-a|< \delta. The "real" proof is the other way, but as long as you are sure all of yours steps are reversible you don't need to state them.

A "real" proof would start, "Given \epsilon, choose \delta= \epsilon. The |x- 4|&lt; \delta= \epsilon so that -\epsilon&lt; x- 4&lt; \epsilon. -\epsilon&amp;lt; (9- x)- 5&amp;lt; \epsilon, |(9- x)- 5|= |f(x)- 5|&amp;lt; \epsilon <br /> <br /> But, as I said, as long as it is clear that all of your steps, in working from \epsilon to delta, are reversible, you don&#039;t need to work from \delta to \delta. What you wrote is perfectly valid.

 

[Saladsamurai]

Keep in mind that you want to show that given any \epsilon &gt; 0 you can find a \delta that works.

You've done the hard part. Your argument is what might be called an exploratory argument. You started out to see how to get |(9x-5)-4 | &lt; \epsilon and wound up with | x-4| &lt; \epsilon.

Now to prove the limit statement you wish to prove that given \epsilon &gt; 0 you can find \delta &gt; 0 such that if 0 &lt; |x-4| &lt; \delta then
|(9x-5)-4 | &lt; \epsilon

So start your final writeup like this:

Suppose \epsilon &gt; 0. Let \delta = \epsilon. (You know this is going to work from your exploratory argument, but you don't need to include that here.) Now you proceed: If 0 &lt; |x-4| &lt; \delta then...now write your exploratory argument in reverse and end up with |(9x-5)-4 | &lt; \epsilon. That shows your \delta works and you are done.

`
That is what is sometimes referred to as "synthetic proof". You have gone from what you want to prove, |f(x)- L|&lt; \epsilon, to your hypothesis, |x-a|&lt; \delta. The "real" proof is the other way, but as long as you are sure all of yours steps are reversible you don't need to state them.

A "real" proof would start, "Given \epsilon, choose \delta= \epsilon. The |x- 4|&lt; \delta= \epsilon so that -\epsilon&lt; x- 4&lt; \epsilon. -\epsilon&amp;lt; (9- x)- 5&amp;lt; \epsilon, |(9- x)- 5|= |f(x)- 5|&amp;lt; \epsilon <br /> <br /> But, as I said, as long as it is clear that all of your steps, in working from \epsilon to delta, are reversible, you don&#039;t need to work from \delta to \delta. What you wrote is perfectly valid.

<br /> <br /> OK great <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" /> Thanks for the pointers. Just another quick question:<br /> <br /> When starting the right way, you both said to &quot;choose epsilon=delta.&quot;<br /> <br /> Assume that I didn&#039;t already know that was the case. I would still make that assumption right? And if it was not the case that epsilon=delta ... what would happen?<br /> <br /> I get the feeling I would just end up with delta being some function of epsilon (or vice versa).<br /> <br /> Thanks <img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f642.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":smile:" title="Smile :smile:" data-smilie="1"data-shortname=":smile:" /><br /> <br /> Casey

 

[Office_Shredder]

Assume that I didn't already know that was the case. I would still make that assumption right? And if it was not the case that epsilon=delta ... what would happen?

In general, epsilon=delta won't suffice (it works here basically because your function is linear with slope 1, so if you make a small change of delta in x, you've made the exact same change in y). Instead, by doing the process of starting with |f(x)-potential limit|<epsilon, you can "solve" the inequality to see what x needs to be. As an example, show that

\lim_{x \rightarrow 0}(2x+1)=1

Without using any algebra of limits stuff. You'll see immediately that epsilon=delta won't suffice

 

[Elucidus]

In general δ is dependent on ε and L. Also if δ suffices then any δ' where 0 < δ' < δ also suffices.

The synthetic process mentioned earlier is usually how δ is determined and once it is found the formal proof begins with "Assume ε > 0 is given, let δ = ..." and the impliction

0 &lt; \left| x - a \right| &lt; \delta \Rightarrow \left| f(x) - L \right| &lt; \epsilon

is shown.

The synthetic part is scratchwork (usually) and is frequently omitted, leading to a lot of "how'd they get that?" moments.


--Elucidus

 

[LCKurtz]

OK great Thanks for the pointers. Just another quick question:

When starting the right way, you both said to "choose epsilon=delta."

Assume that I didn't already know that was the case. I would still make that assumption right? And if it was not the case that epsilon=delta ... what would happen?

I get the feeling I would just end up with delta being some function of epsilon (or vice versa).

Thanks

Casey

You don't "choose epsilon=delta.". epsilon is given and you choose delta = epsilon (in this problem). Delta will usually depend on epsilon and the point of your exploratory argument is to figure out a function of epsilon to set delta to make it work.

 

[Saladsamurai]

In general δ is dependent on ε and L. Also if δ suffices then any δ' where 0 < δ' < δ also suffices.

The synthetic process mentioned earlier is usually how δ is determined and once it is found the formal proof begins with "Assume ε > 0 is given, let δ = ..." and the impliction

...
...
The synthetic part is scratchwork (usually) and is frequently omitted, leading to a lot of "how'd they get that?" moments.


--Elucidus

Ah, I see

That is a little frustrating. Thank you.