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(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}x^{2}=c^{2}.

Solution:

Consider

|x^{2}- c^{2}| < 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] |x^{2}- c^{2}| < 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!

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# Referring to proved lim of the function.

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