- #1

- 95

- 0

Please use laymans english to explain, thank you

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter j-lee00
- Start date

- #1

- 95

- 0

Please use laymans english to explain, thank you

- #2

mjsd

Homework Helper

- 726

- 3

what are you asking?

- #3

- 95

- 0

how the proof works using the formal definition of a limit

- #4

mjsd

Homework Helper

- 726

- 3

with epsilon and delta?

- #5

- 95

- 0

yep thats correct

- #6

Gib Z

Homework Helper

- 3,346

- 5

The proof works by showing that in the close neighborhood of the value that the limit is approaching, all the values in the neighborhood as becoming the same.

- #7

mjsd

Homework Helper

- 726

- 3

suppose we want to show

[tex]\lim_{x\rightarrow x_0} f(x) = L[/tex],

then we would like to know whether f(x) indeed gets close to L as x gets close to x_0. And that is exactly what the definition is doing.

now in proofs, you often wanna find a relationship between [tex]\delta[/tex] and [tex]\epsilon[/tex]; and often what that does is to simply ensure that we can ALWAYS find the appropriate values for them so that the the inequalities involving [tex]\delta[/tex] and [tex]\epsilon[/tex] can be satisfied.

- #8

- 95

- 0

thank you very much

- #9

HallsofIvy

Science Advisor

Homework Helper

- 41,833

- 963

To prove that [itex]lim_{x\rightarrow 3} 2x+ 1= 7[/itex] a typical argument goes "If [itex]|f(x)-L|= |2x+1- 7|< \epsilon[/itex] then [itex]\|2x- 6|= 2|x-3|< \epsilon[/itex] so it suffices to take [itex]|x-3|< \delta= \epsilon/2[/itex]".

That seems "circular" because we start from [itex]|f(x)-L|< \epsilon[/itex] which, by the definition of limit, is what we want to show. You are correct that that is

"Given [itex]\epsilon[/itex], take [itex]\delta= \epsilon/2[/itex] (which looks like the professor is picking it out of the air since the student didn't see the analysis above). Then if [itex]0< |x-3|< \delta= \epsilon/2[/itex], we have [itex]0< 2|x-3|= |2x- 6|= |2x+ 1- 7|< \epsilon[/itex]".

Typically, having done the "analysis" we don't need to write out the "true" proof because it is clear that every step in going from [itex]\epsilon[/itex] to [itex]\delta[/itex] is

- #10

mjsd

Homework Helper

- 726

- 3

Share:

- Replies
- 19

- Views
- 3K