Uniform convergence of sequence of function necessary for switching limits

Question

If a sequence of function $f_n(x)$ converges to $f(x)$ uniformly, we know that we have $$\lim_{t\to x}\lim_{n\to\infty}f_n(t)=\lim_{n\to\infty}\lim_{t\to x}f_n(t)$$

My question is: Is uniformly convergence necessary for the above to hold?

Answer 1

Suppose on $[0,1]$ we set $f_n(t) = nt(1-t)^n.$ Then $f_n(t) \to 0$ pointwise on $[0,1].$ Let $x=0.$ Then

$$\lim_{t\to x}\lim_{n\to\infty}f_n(t)=\lim_{n\to\infty}\lim_{t\to x}f_n(t).$$

But since $f_n(1/n) = (1-1/n)^n \to 1/e, \,f_n$ fails to converge uniformly to $0$ in every $[0,\delta].$

Answer 2

Uniform convergence is not necessary, consider e.g. $f_n(t)=t^n$ and $x=0$.

However, if $f_n$ does not converge uniformly to $f$, there may still be some $x$, for which the equality fails to hold, e.g. same example as above with $x=1$.