Question

Analysis question.

Answer 1

That's the question and this is what i got. I would like to know if it's right. It might take me a while to write it.

Answer 2

\[\forall x_{0} \in \mathbb{R} \forall \epsilon >0\exists \delta >0\forall x \in \mathbb{R}\]

Answer 3

\[\Delta f = f(x+\Delta x) - f(x) = ( x + \Delta x)^{2} -x ^{2} = 2x \Delta x + \Delta x ^{2}\]
Function is continuous if it's increment gose to 0 as increment of x is going to 0. Just take a lim as x--->0 of the expresion above and you got your proof

Answer 4

\[\left[ \left| x-x _{0} \right|<\epsilon \rightarrow \left| x^{2} -x _{0}^{2} \right| <\epsilon\right]\]

Answer 5

choose \[x_{0}\]
Let \[a=x_{0}+1 and \delta=\min \left( 1, \epsilon/2a \right)\]

Answer 6

right, just \[|x-x _{0}|< \delta \rightarrow |x-x _{0}|^{2}<\epsilon\]

Answer 7

okay i think what i have is right then.

Answer 8

thanks

Answer 9

you welcome