Question

negate the statement: for every epsilon>0, there exists delta>0, such that abs(f(x)-f(0)) 11 years ago

Answer 1

to negate 'for all x , P(x) ' use ' there exists x such that not P(x) '

Answer 2

i know, but im having trouble since there is a 'for all' and 'there exists' in the original statement

Answer 3

Do you mean
negate the statement: for every epsilon>0, there exists delta>0, such that abs(f(x)-f(0))<epsilon for all -delta<x<delta

Answer 4

yes, i did

Answer 5

You put 0 instead of x between -delta < x < delta

Answer 6

\[
\text { There is } \epsilon > 0 \text { such that, for very } \delta >0, \text { there is an }
x ; \quad -\delta < x < \delta \\
\text { with } |f(x) - f(0)| > \epsilon
\]

Answer 7

every instead of very

Answer 8

let me rewrite the original statement , so its clearer

Answer 9

for every epsilon>0, there exists delta>0, such that *if* -delta<x<delta then abs(f(x)-f(0))<epsilon

Answer 10

thank you!

Answer 11

to negate 'if p then q ' , use p and not q

Answer 12

also you could write the original statement as
for every epsilon>0, there exists delta>0, such that abs(f(x)-f(0))<epsilon *whenever* -delta<x<delta

Answer 13

for every epsilon>0, there exists delta>0, such that *if* -delta<x<delta then abs(f(x)-f(0))<epsilon

negation: there exists e >0 such that for all d > 0 , -d < x < d and | f (x) - f(0) |> e