- #1

agro

- 46

- 0

I think I understand the formal/rigorous definition of limit, but I find proving various limits (or following proofs of them) extremely difficult. I hope you all will help me. I think I won't advance my calculus study until I really get this limit proving thing btw...

Here's one limit prove that baffles me...

Prove that lim

We must show that given any [ee] > 0 there exist [pard] > 0 such that

|x

OK here...

|x + 3| |x - 3| < [ee] if 0 < |x - 3| < [pard] (2)

Using triangle inequality, we see that

|x + 3| = |(x - 3) + 6| [<=] |x - 3| + 6

No problem here...

Therefore if 0 < |x - 3| < [pard]

|x + 3| |x - 3| [<=] (|x - 3| + 6) |x - 3| < ([pard] + 6)[pard]

Fine... Now it's getting more difficult

It follows that (2) will be satisfied for any positive value of [pard] such that ([pard] + 6)[pard] [<=] [ee]. Let us agree to restrict our attention to positive values of [pard] such that [pard] [<=] 1. With this restriction, ([pard] + 6)[pard] [<=] 7[pard], so that (2) will be satisfied as long as it is also the case that 7[pard] [<=] [ee].

Understandable...

We can achieve this by taking [pard] to be the minimum of the numbers [ee]/7 and 1.

Whoa... Now I didn't really understand this last part... The writer just states it like magic (to me)... Anyone can give a more detailed explanation? I attached an image relating to my understanding btw....

Thanks a lot (other limit question will follow).

PS: I also found it is possible (and more straightforward for me) to prove the existence of the limit by solving ([pard] + 6)[pard] [<=] [ee] using the abc formula...

Here's one limit prove that baffles me...

Prove that lim

_{x->3}x^{2}= 9We must show that given any [ee] > 0 there exist [pard] > 0 such that

|x

^{2}- 9| < [ee] if 0 < |x - 3| < [pard] (1)OK here...

|x + 3| |x - 3| < [ee] if 0 < |x - 3| < [pard] (2)

Using triangle inequality, we see that

|x + 3| = |(x - 3) + 6| [<=] |x - 3| + 6

No problem here...

Therefore if 0 < |x - 3| < [pard]

|x + 3| |x - 3| [<=] (|x - 3| + 6) |x - 3| < ([pard] + 6)[pard]

Fine... Now it's getting more difficult

It follows that (2) will be satisfied for any positive value of [pard] such that ([pard] + 6)[pard] [<=] [ee]. Let us agree to restrict our attention to positive values of [pard] such that [pard] [<=] 1. With this restriction, ([pard] + 6)[pard] [<=] 7[pard], so that (2) will be satisfied as long as it is also the case that 7[pard] [<=] [ee].

Understandable...

We can achieve this by taking [pard] to be the minimum of the numbers [ee]/7 and 1.

Whoa... Now I didn't really understand this last part... The writer just states it like magic (to me)... Anyone can give a more detailed explanation? I attached an image relating to my understanding btw....

Thanks a lot (other limit question will follow).

PS: I also found it is possible (and more straightforward for me) to prove the existence of the limit by solving ([pard] + 6)[pard] [<=] [ee] using the abc formula...

#### Attachments

Last edited: