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[QUOTE="anuttarasammyak, post: 6821290, member: 674023"] [spoiler] [tex]\forall \epsilon>0 \ \exists \delta> 0 \ \ |f(x+\delta)-f(x)|<\epsilon[/tex] with no reard to individual value of x. Summing up this in x interval of [0,x] x>0 [tex]|f(x)-f(0)|\leq \frac{\epsilon}{\delta}|x|+\epsilon[/tex] we get same formula for [x,0] x<0. So for any x [tex]|f(x)| \leq \frac{\epsilon}{\delta}|x|+|f(0)|+\epsilon \leq A+B|x|[/tex] where [tex]|f(0)|+\epsilon \leq A , \frac{\epsilon}{\delta}\leq B[/tex] [EDIT] Thanks to the sugegstion in post #13, I would add a sentense to the second line : with no reard to individual value of x. [B]Let us fix [/B]##\epsilon##[B] and [/B]##\delta##[B] which satisfy the inequality.[/B] Summing up this in x interval of [0,x] x>0 [/spoiler] [/QUOTE]
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