- #1
A Dhingra
- 211
- 1
hello..
(My professor solved this question in class but i have a few doubts in it,and this is probably the method used by Sherbert and Beartle in intro to real analysis.)
Prove that lim x→c x2 =c2.
Solution:
Consider
|x2 - c2| < ε --1)
|x-c||x+c|< ε
Suppose
| x-c|<1
so, |x| ≤ |c| +1
|x+c|≤ 2|c|+1
so,
|x-c||x+c| ≤ (2|c|+1)|x-c|< ε --2)
then,
|x-c| < ε/(2|c| +1)
hence
δ = inf{1, ε/(2|c|+1)}
Hence
|x-c| <δ [itex]\Rightarrow[/itex] |x2 - c2| < ε
(This was so far as done in class.)
Now here we assumed the first statement, while this doesn't necessarily mean 2)...
We know, |x-c||x+c|< ε
then putting a constraint on |x-c|, then
finding its max(|x+c|)
How do we show that the product of these two is still less then ε.
Next thing i could not understand was, for a specific value of ε, there exists an upper limit for the value of δ, then how can we randomly select δ to be one and then revise the value of δ ?
Thanks for any help!
(My professor solved this question in class but i have a few doubts in it,and this is probably the method used by Sherbert and Beartle in intro to real analysis.)
Prove that lim x→c x2 =c2.
Solution:
Consider
|x2 - c2| < ε --1)
|x-c||x+c|< ε
Suppose
| x-c|<1
so, |x| ≤ |c| +1
|x+c|≤ 2|c|+1
so,
|x-c||x+c| ≤ (2|c|+1)|x-c|< ε --2)
then,
|x-c| < ε/(2|c| +1)
hence
δ = inf{1, ε/(2|c|+1)}
Hence
|x-c| <δ [itex]\Rightarrow[/itex] |x2 - c2| < ε
(This was so far as done in class.)
Now here we assumed the first statement, while this doesn't necessarily mean 2)...
We know, |x-c||x+c|< ε
then putting a constraint on |x-c|, then
finding its max(|x+c|)
How do we show that the product of these two is still less then ε.
Next thing i could not understand was, for a specific value of ε, there exists an upper limit for the value of δ, then how can we randomly select δ to be one and then revise the value of δ ?
Thanks for any help!