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Main Question or Discussion Point
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_{x>3} x^{2} = 9
We 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...
Here's one limit prove that baffles me...
Prove that lim_{x>3} x^{2} = 9
We 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...
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