Change of order of double limit of function sequence The more general quesion is under what conditions the folloing equality will hold (all functions are real valued):
$$\lim_{x \rightarrow a} \ \lim_{j \rightarrow \infty} f_j(x) = \lim_{j \rightarrow \infty} \ \lim_{x \rightarrow a} f_j(x)$$
A more specific question is if it will hold for non-continuous functions $f_j$ that are uniformly convergent to a non-continuous limit function $f$ and all the single and iterated double limits exist.
Also some references would be useful.
 A: Uniform convergence means that for every $\varepsilon > 0$ there exists $j_0$ such that for all $j \geq j_0$ and for all $x \in D$(the domain of definition for the functions $(f_j),f$ you have $|f_j(x)-f(x)| <\varepsilon$. 
By the way the question is defined, although $f,f_j$ are not continuous, the limits $L=\lim_{x \to a}f(x)$ and $L_j=\lim_{x \to a}f_j(x)$ exist. The question translates in proving or disproving the followint equailty:

$$\lim_{j \to \infty}L_j=L$$

Let $\varepsilon>0$, and from the definition of uniform continuity we know that there exists $j_0$ such that forall $j \geq j_0$ and forall $x \in D$ we have that $|f_j(x)-f(x)|<\varepsilon$. Taking $x \to a$ in the last inequality we get that $|L_j-L|<\varepsilon ,\ \forall j \geq j_0$. This proves the assertion.
As a remark, I think the question should be edited such that


*

*the domain of definition of the functions $f_j,f$ is clear, i.e. $f_j,f: D \to \Bbb{R}$ where $D=\Bbb{R}$ or some other suitable set. 

*mention that $a$ is a limit point for $D$

*although the functions $f_j,f$ are not continuous, for the question to be valid, we must assume the existence of the limits $\lim_{x \to a}f_j(x), \ \lim_{x \to a}f(x)$.
A: Note that 
$$\left |\lim_{x\to a}f_j(x) - \lim_{x\to a} f(x)\right|\le \|f_j-f\|_\infty\to 0,\; j\to \infty$$
Thus $$\lim_{j\to\infty}\lim_{x\to a}f_j(x) = \lim_{x\to a} f(x) =  \lim_{x\to a} \lim_{j\to\infty}f(x)$$
A: The answer to the specific question is Yes.
Since $f_j$ converges (uniformly and hence) pointwise to $f$,
$$
\mathop {\lim }\limits_{x \to a} \mathop {\lim }\limits_{j \to \infty } f_j (x) = \mathop {\lim }\limits_{x \to a} f(x).
$$
For any $\varepsilon > 0$, since $f_j$ converges uniformly to $f$, there exists $N = N(\varepsilon)$ such that
$$
\sup _x |f_j (x) - f(x)| < \varepsilon 
$$
for any $j > N$. We assume that all limits exist. Hence, for any $j > N$,
$$
|\lim _{x \to a} f_j (x) - \lim _{x \to a} f(x)| = |\lim _{x \to a} (f_j (x) - f(x))| \le \varepsilon .
$$
Define $p_j = \lim _{x \to a} f_j (x)$ and $p = \lim _{x \to a} f(x)$. Then,
$$
|\lim _{j \to \infty } p_j  - p| = \lim _{j \to \infty } |p_j  - p| \le \varepsilon ,
$$
since $|p_j - p| \leq \varepsilon$ for any $j > N$. Since $\varepsilon$ is arbitrary, $
\lim _{j \to \infty } p_j  = p$, hence
$$
\mathop {\lim }\limits_{j \to \infty } \mathop {\lim }\limits_{x \to a} f_j (x) = \mathop {\lim }\limits_{x \to a} f(x) = \mathop {\lim }\limits_{x \to a} \mathop {\lim }\limits_{j \to \infty } f_j (x).
$$
