Why is it fine to assume .5 | x - 4 | < e in an epsilon-delta proof?

[bjgawp]

Hey there everyone. I was looking at an epsilon-delta proof I did and realized that I wasn't exactly sure why one of my statements was true:

http://img255.imageshack.us/img255/7356/proofmy5.jpg

On the third last line, why is it fine to assume that .5 | x - 4 | < e is true? Isn't there a chance that .5 | x - 4 | may be greater than e?

Thanks in advance!

Edit: Hmm, wonder why the IMG tags don't work ...

 

[JeffN]

it's not okay to assume that.The inequality \left|\sqrt{x}-2\right|\leq \frac{1}{2} \left|\ x - 4 \right|\ shows that if you restrict the x-values so that \left|\ x - 4\right|\ &lt;2\epsilon for the given \epsilon, then you get

\left|\sqrt{x}-2\right|\leq \frac{1}{2} \left|\ x - 4 \right|\ &lt; \frac{1}{2} (2\epsilon) =\epsilon

Also, the last line, 'And since . . .' is a bit weird. Remember that you're showing that such a delta exists in the first place.

 

[bjgawp]

How does \left|\sqrt{x}-2\right|\leq \frac{1}{2} \left|\ x - 4 \right|\ show that \left|\ x - 4\right|\ &lt;2\epsilon without assuming \frac{1}{2} \left|\ x - 4 \right|\ &lt; \epsilon first? Sorry. I don't see how the first inequality connects with it being less than epsilon. Thanks for the help!

Oh and yeah, I typed this up haphazardly without thinking what I meant by the last line. Thanks again!

 

[JeffN]

remember that you fixed \epsilon as a positive number.

the inequality shows that IF you make \left|\ x - 4\right|\ &lt; \epsilon , THEN you get \left|\sqrt{x}-2\right|\&lt;\epsilon.

Read over the definition of a limit carefully and see how it applies to this particular problem:

We say that \lim_{x\to\\a}f(x)=v whenever,

for all \epsilon &gt; 0, there is some \delta &gt;0 such that,

whenever we have 0&lt;\left|\ x-a \right|\ &lt; \delta, it follows that \left|\ f(x) - v\right|\ &lt; \epsilon

 

[HallsofIvy]

The reason that last line seems a bit "peculiar" is that it is really a peculiarity of the way we do proofs of limits.

The definition of limit requires that we show that, for a specific \delta, if |x-a|&lt; \delta, then |f(x)- L|&lt; \epsilon. But what we do is work the other way- assuming a value of \epsilon, we calculate the necessary \delta. The point is that every step is "reversible"- you could start from, in this case, |x-4|< \delta and, by just going through the derivation "in reverse" arrive at |\sqrt{x}- 2|&lt; \epsilon.

That's sometimes referred to as "synthetic" proof. Again, we go from the conclusion we want to the hypothesis- but it is crucial that every step be "reversible"!