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

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[SOLVED] Help Deciphering limit text

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

How is the book allowed to set [tex]3|x-2|<3\delta[/tex](the 3 * delta), and better yet,
[tex]3|x-2|=\epsilon[/tex](= 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 [tex]3|x-2|<\epsilon[/tex] rather than [tex]3|x-2|=\epsilon[/tex]

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

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

 

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

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

Thanks a bunch.

 

[HallsofIvy]

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

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

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

, and better yet,
[tex]3|x-2|=\epsilon[/tex](= epsilon)?

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

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

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

 

[razored]

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


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

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 [itex]\delta[/itex], not 3|x-2|.




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

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

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