Introduction to Proofs / Logic and setsNegating an ε-δ statementChooseQuantifiers and negationWhich statement is the negation of ∀ε>0 ∃δ>0 ∀x: ∣x∣<δ⇒∣f(x)∣<ε ?\forall \varepsilon > 0\ \exists \delta > 0\ \forall x:\ |x| < \delta \Rightarrow |f(x)| < \varepsilon \,?∀ε>0 ∃δ>0 ∀x: ∣x∣<δ⇒∣f(x)∣<ε? (The original statement says that f(x)→0f(x) \to 0f(x)→0 as x→0x \to 0x→0.)Choose one answer∃ε>0 ∀δ>0 ∃x: ∣x∣<δ and ∣f(x)∣≥ε\exists \varepsilon > 0\ \forall \delta > 0\ \exists x:\ |x| < \delta \text{ and } |f(x)| \ge \varepsilon∃ε>0 ∀δ>0 ∃x: ∣x∣<δ and ∣f(x)∣≥ε∃ε>0 ∀δ>0 ∀x: ∣x∣<δ⇒∣f(x)∣≥ε\exists \varepsilon > 0\ \forall \delta > 0\ \forall x:\ |x| < \delta \Rightarrow |f(x)| \ge \varepsilon∃ε>0 ∀δ>0 ∀x: ∣x∣<δ⇒∣f(x)∣≥ε∀ε>0 ∃δ>0 ∃x: ∣x∣≥δ and ∣f(x)∣<ε\forall \varepsilon > 0\ \exists \delta > 0\ \exists x:\ |x| \ge \delta \text{ and } |f(x)| < \varepsilon∀ε>0 ∃δ>0 ∃x: ∣x∣≥δ and ∣f(x)∣<ε∃ε≤0 ∀δ≤0 ∃x: ∣x∣<δ and ∣f(x)∣≥ε\exists \varepsilon \le 0\ \forall \delta \le 0\ \exists x:\ |x| < \delta \text{ and } |f(x)| \ge \varepsilon∃ε≤0 ∀δ≤0 ∃x: ∣x∣<δ and ∣f(x)∣≥εCheck answerHint 1 of 2