Why can we assume 0.5|x - 4| < ε in epsilon-delta proofs?

[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|\ <2\epsilon$$ for the given $$\epsilon$$, then you get

$$\left|\sqrt{x}-2\right|\leq \frac{1}{2} \left|\ x - 4 \right|\ < \frac{1}{2} (2\epsilon) =\epsilon$$

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|\ <2\epsilon$$ without assuming $$\frac{1}{2} \left|\ x - 4 \right|\ < \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 $$\left|\ x - 4\right|\ < \epsilon$$ , THEN you get $$\left|\sqrt{x}-2\right|\<\epsilon$$.

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 $$\epsilon > 0$$, there is some $$\delta >0$$ such that,

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

 

[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 $|x-a|< \delta$, then $|f(x)- L|< \epsilon$. 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 $|\sqrt{x}- 2|< \epsilon$.

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"!