Negate the following sentence: $\forall x, |x−a| < δ \Rightarrow |f(x)−L| < \varepsilon$ Negate the sentence: $\forall x, |x−a| < δ \Rightarrow |f(x)−L| < \varepsilon$ For my negation I got: $\exists x, |x-a| < δ \Rightarrow |f(x)-L| \geq \varepsilon$ Would that be correct?

You are correct that $\lnot(\forall x P(x))$ is $\exists x \lnot P(x)$, but your negation of the implication is incorrect. The negation of $P \implies Q$ is $P \land \lnot Q$. See for example this truth table:


P Q P ⇒ ¬Q P ∧ ¬Q
============================
T T F F
T F T T
F T T F
F F T F


In this case, the negation should read in plain English as something like "there is an $x$ such that $x$ is $\delta$-close to $a$ and $f(x)$ is at least $\varepsilon$ away from $L$."