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!