Precise Definition of a Limit

[Chris*]

I'm having a difficult time with trying to understand the whole epsilon-delta concept.
Here's my homework problem:

Suppose that for a given ? and ? , it is known that |f(x) - 3|<? if 0<|x - 4|<?.
Which of the following must also be true? Explain.

\(\displaystyle 1)\;\left|f(x)-3 \right|<\frac{\epsilon}{2}\quad \mathrm{if}\quad 0<\left|x-4 \right|<\frac{\delta}{2}\)
\(\displaystyle 2)\;\left|f(x)-3 \right|<\epsilon\quad \mathrm{if}\quad 0<\left|x-4 \right|<\frac{\delta}{2}\)
\(\displaystyle 3)\;\left|f(x)-3 \right|<2\epsilon\quad \mathrm{if}\quad 0<\left|x-4 \right|<\delta\)
\(\displaystyle 4)\;\left|f(x)-3 \right|<\epsilon\quad \mathrm{if}\quad 0<\left|x-4 \right|<2\delta\)

I think 2 is true because you will still get an ? > |f(x)-3| if you choose ?/2. It is smaller than ? so that means it still would fit into the range of values ? for the original statement.

I also think 3 is true. In this case, ? is doubled. So that would extend the range of allowable values of ?. So, if ? is the same it shouldn't exceed these values and you should get a number within L-2? and L+2?.

I don't think 1 and 4 is correct because in those cases ? may exceed the boundaries for ?.

Do my explanations make sense? Or am I looking at this problem incorrectly?

Any help is appreciated!

 

[BigGlenntheHeavy]

For 1, false as |f(x)-3| < ?, ? could be > than ?/2 when |x-4|< ?/2.

2 is true

3 and 4 are false, as 3 exceeds its range and 4 exceeds its domain.

Note: This is how I see it, I could be wrong.

 

[pka]

Suppose that for a given ? and ? , it is known that |f(x) - 3|<? if 0<|x - 4|<?.
Which of the following must also be true? Explain.
\(\displaystyle 3)\;\left|f(x)-3 \right|<2\epsilon\quad \mathrm{if}\quad 0<\left|x-4 \right|<\delta\)
I also think 3 is true. In this case, ? is doubled. So that would extend the range of allowable values of ?. So, if ? is the same it shouldn't exceed these values and you should get a number within L-2? and L+2?.

That is correct. Because \(\displaystyle \epsilon < 2\epsilon\)