Sequence of Lipschitz maps that converge pointwise to a Lipscthiz functions must have bounded Lipschitz constants?

Question

Let $f_n:M \to M$ be a sequence of Lipschitz maps in a metric space $(M,d)$. Assume that we know that $\{f_n\}_{n=1}^\infty$ converge pointwise to $f:M\to M$ that is also a Lipschitz map. Let $L_n$ denote the Lipschitz constant of each $f_n$, that is, $L_n=\sup_{x\neq y}\dfrac{d(f(x),f(y))}{d(x,y)}$. Then, can we deduce that the sequence $\{L_n\}_{n=1}^\infty$ is bounded?

I expect this being false, but I can't find a counterexample. Maybe there is hope to argue that converging pointwise to a Lipscthiz map $f$ must bound the Lipschitz constants of the $\{f_n\}$ sequence (or otherwise we would have a contradiction)?

Answer 1

$f_n(x)=\frac 1n\sin(n^2x)$, $f(x)=0$. Then $f_n\to f$ uniformly. All the functions are Lipschitz. However $f_n'(x)=n\cos(n^2x)$ so $L_n=n$.