If $f_n \to f$ uniformly on $[0,1]$ and $f$ is Lipschitz-continuous, each $f_n$ is necessarily Lipschitz too?

Question

Let $\{ f_n \}$ be a sequence of continuous functions on $[0,1]$ converging uniformly to some $f$.

Suppose further that $f$ is Lipschitz-continuous on $[0,1]$. Then, does there exists some $N \in \mathbb{N}$ such that each $f_n$ is necessarily Lipschitz for $n \geq N$?

I suspect this is the case, but cannot prove / disprove this statement.. Could anyone help me?

Answer 1

Let $f_n$ be given by $1/n \sqrt x +(n-1)/n$. None of these functions is Lipschitz, but the uniform limit is constant.