How Can I Use Delta and Epsilon to Understand Limits in Calculus?

[razored]

[SOLVED] Help Deciphering limit text

"The function f(x) = 3x aproaches the limit 6 as x\rightarrow 2. In fact, given any \epsilon > 0, choose \delta = \frac {\epsilon} {3}. We then have
|f(x)-6|=|3x-6|=3|x-2|<3\delta = \epsilon whenever 0<|x-2|<\delta."

How is the book allowed to set 3|x-2|<3\delta(the 3 * delta), and better yet,
3|x-2|=\epsilon(= epsilon)? I thought the limit designated to them was delta alone, not 3 delta. In addition, how are they allowed to set it equal to epsilon?

I am really trying to understand this; I realize the answer to my questions may seem, "elementary", but I am trying to master the basics of Calculus(currently in pre-calculus).

 

[konthelion]

It is 3|x-2|<\epsilon rather than 3|x-2|=\epsilon

By definition of the \epsilon-\delta criterion of a point.
\forall \epsilon>0,\exists \delta>0
|f(x)-f(x_{0})|<\epsilon if |x-x_{0}|<\delta

The author of that text used \delta = \frac{\epsilon}{3}, substitute that back into
3|x-2|<\delta. The trick is that we must find a \delta>0 such that |3x-6| < \epsilon

 

[razored]

|x-x_{0}|< \delta The definition
|3x-6|< \delta substitution
3|x-2|<\delta Factoring
I know the author substituted, but if he substituted \delta = \frac {\epsilon}{3} into the equation, he would get 3|x-2|<\frac {\epsilon}{3}; not 3|x-2|<3\delta. I'm not catching on.

Also, you mentioned "The trick is that we must find a \delta>0 such that |3x-6|<\epsilon." Another question, how do we know what we are actually looking for, the epsilon or the delta?

Thanks a bunch.

 

[HallsofIvy]

"The function f(x) = 3x aproaches the limit 6 as x\rightarrow 2. In fact, given any \epsilon > 0, choose \delta = \frac {\epsilon} {3}. We then have
|f(x)-6|=|3x-6|=3|x-2|<3\delta = \epsilon whenever 0<|x-2|<\delta."

How is the book allowed to set 3|x-2|<3\delta(the 3 * delta)

It didn't "set" it. It is saying that if |x-2|< \delta then 3|x-2|< 3\delta, multiplying both sides of the inequality by 3.

, and better yet,
3|x-2|=\epsilon(= epsilon)?

It isn't and I'll bet the book doesn't say that. Starting from |x-2|< \delta, multiply both sides of the inequality by 3 to get 3|x-2|< 3\delta. Now, since \delta is defined to be \epsilon/3, it follows that 3\delta= \epsilon so 3|x- 2|= |3x- 6|< \epsilon.

I thought the limit designated to them was delta alone, not 3 delta. In addition, how are they allowed to set it equal to epsilon?[/itex]

Designated what to be "delta alone"? |x- 2| is given as less than \delta, not 3|x-2|.

I am really trying to understand this; I realize the answer to my questions may seem, "elementary", but I am trying to master the basics of Calculus(currently in pre-calculus).

Also, you mentioned "The trick is that we must find a such that ." Another question, how do we know what we are actually looking for, the epsilon or the delta?

The definition of "\lim_{x\rightarrow a} f(x)= L" is "Given \epsilon> 0 , there exist \delta>0 such that if 0< |x-a|< \delta, then |f(x)- L|< \epsilon".

To "prove" that a limit is correct, you have to show that definition is true. You are given epsilon, it could be any number. You need to prove that such a \delta exists and a very good way of showing something exists is to find it. "Given" \epsilon, you need to find (and so are "looking for") delta.

 

[razored]

It didn't "set" it. It is saying that if |x-2|< \delta then 3|x-2|< 3\delta, multiplying both sides of the inequality by 3.


It isn't and I'll bet the book doesn't say that. Starting from |x-2|< \delta, multiply both sides of the inequality by 3 to get 3|x-2|< 3\delta. Now, since \delta is defined to be \epsilon/3, it follows that 3\delta= \epsilon so 3|x- 2|= |3x- 6|< \epsilon.

I thought the limit designated to them was delta alone, not 3 delta. In addition, how are they allowed to set it equal to epsilon?[/itex]
Designated what to be "delta alone"? |x- 2| is given as less than \delta, not 3|x-2|.




The definition of "\lim_{x\rightarrow a} f(x)= L" is "Given \epsilon> 0 , there exist \delta>0 such that if 0< |x-a|< \delta, then |f(x)- L|< \epsilon".

To "prove" that a limit is correct, you have to show that definition is true. You are given epsilon, it could be any number. You need to prove that such a \delta exists and a very good way of showing something exists is to find it. "Given" \epsilon, you need to find (and so are "looking for") delta.

Ok, that clears up most of my confusion. Thanks again Halls of Ivy!