# Homework Help: Prove the Limit Statement

1. Sep 15, 2009

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.

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

2. Sep 15, 2009

### VeeEight

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

3. Sep 15, 2009

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.

4. Sep 15, 2009

### 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.

5. Sep 15, 2009

### HallsofIvy

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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 $|x- 4|< \delta= \epsilon$ so that $-\epsilon< x- 4< \epsilon. [itex]-\epsilon< (9- x)- 5< \epsilon$, $|(9- x)- 5|= |f(x)- 5|< \epsilon$

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't need to work from $\delta$ to $\delta$. What you wrote is perfectly valid.

6. Sep 15, 2009

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

7. Sep 15, 2009

### Office_Shredder

Staff Emeritus
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

8. Sep 15, 2009

### 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

$$0 < \left| x - a \right| < \delta \Rightarrow \left| f(x) - L \right| < \epsilon$$

is shown.

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

--Elucidus

9. Sep 15, 2009

### LCKurtz

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.

10. Sep 16, 2009