Why if $\lim_{n \rightarrow \infty} f_n(x)=f(x)$ for all $x \in X$, we cannot directly say that $f(x)$ is the uniform limit of $f_n(x)$ as well? I'm trying to understand why if, given $f_n(x)$ with $x \in X$ and $\lim_{n \rightarrow \infty} f_n(x)=f(x)$ for all $x \in X$, we cannot directly say that $f(x)$ is the uniform limit of $f_n(x)$. Can you confirm me that the reason is that according to the definition of pointwise convergence:
$\forall \epsilon > 0 \forall x\in X  \exists N\in\mathbb{N} \forall n\geq N \ \ s.t. \ \ |f_n(x) - f(x)| < \epsilon$
We know that there exists an $N$ for every $x$ BUT we cannot know that this $N$ applies for every $x$? 
 A: The usual counterexample: $f_n(x) = x^n$, $x\in[0,1]$.
A: Yes, exactly. Take the sequence of functions $$
  f_n(x) = x^n
$$
Obviously, on the interval $(0,1)$, $$
  f(x) = \lim_{n\to\infty} f_n = 0.
$$
But OTOH, if you fix an $N$, you can always find an $x$ such that $$
  \left|f(x) - f_N(x)\right| \geq \frac{1}{2}.
$$ Just pick $x = \sqrt[N]{\frac{1}{2}}$, then $$
  \left|1 - \left(\sqrt[N]{\frac{1}{2}}\right)^N\right| = 1 - \frac{1}{2} = \frac{1}{2}.
$$
The convergence can thus not be uniform, because in the case of uniform convergence, it should be possible to choosen an $N$ such that no matter which $x$ you pick, the distance of $f_N(x)$ from the limit $f(x)$ should be less than $\frac{1}{2}$. (And of course there should be a similar picking procedure for $N$ for any other $\epsilon$ you chose, not only for $\frac{1}{2}$).
A: Another simple example is $f_n(x)=x/n,x \in \Bbb R$,  converging pointwise to $f \equiv 0$, but not uniformly.
Let $\epsilon = 1$, for instance. Pick $N \in \Bbb N$. You want to find $x \in \Bbb R$ such that $|f_n(x) - 0|>1$, i.e. $|x|>1 \cdot N$. You can take $x=N + 1$ for instance. Then the convergence of $(f_n)_{n≥1}$ to $f$ is not uniform.
