Pointwise convergence and uniform convergence converge to same limit? Sometimes I see in proofs they say "this function is pointwise convergent to limit a. Let's check if it's uniformly convergent to limit a." And then they do a uniform convergence check on a and prove that it cannot converge uniformly to limit a. 
My question is, if we know a function converges pointwise to a limit, suppose it does converge uniformly, would it converge to the same limit?
 A: Yes, it does. Let $(f_n)$ be a series of functions converging uniformly to $f$. Then $\forall x$ we have $|f_n(x) - f(x)| \leq \sup_x{|f_n(x) - f(x)|}$. Thus $\lim_{n \to \infty}f_n(x) = f(x)$ (pointwise convergence). This limit is unique.
A: I will explain this using definitions of both pointwise convergence and uniform convergence
suppose $ f_n : A\rightarrow\mathbb{R} $
Pointwise convergence definition 
$$
(\forall x \in A) (\forall \varepsilon>0) (\exists n_0=n_0(x,\varepsilon) \in \mathbb{N}) (\forall n>n_0) \left | f(x) - f_n(x) \right |<\varepsilon
$$
Uniform convergence definition
$$
(\forall \varepsilon>0) (\exists n_0=n_0(\varepsilon) \in \mathbb{N}) (\forall x \in A)  (\forall n>n_0) \left | f(x) - f_n(x) \right |<\varepsilon
$$
Note that $ n_0(x,\varepsilon) $ means that $n_0$ is a function of $ x $ and $ \varepsilon $
And $ n_0(\varepsilon) $ is a function of $ \varepsilon $
Suppose $f_n\rightrightarrows f$ on $A$
So since in uniform convergence we can find such $ n_0=n_0(\varepsilon) $ that depends just on $ \varepsilon$ we can use that same $n_0$ for definition on pointwise convergence.
Our chosen $n_0$ works for every $ x $ so for every $ x $ we can chose our $n_0=n_0(\varepsilon)$
That means functions $f_n$ converges poinwise to $f$ on $A$ when $f_n\rightrightarrows f$ on $A$
Reverse does not apply If a function converges it doesn't need to converge uniformly but if it does the limit is the same as pointwise convergence.
A: Suppose there exists some $\{f_n\}$ so that the sequence converges pointwise to $f$, but $\{f_n\}$ converges uniformly to $g \neq f$. Then there exists some $x \in \mathbb{R}$ so that $g(x) \neq f(x)$.
Let $\epsilon=\frac{|g(x)-f(x)|}{2}$...
This boils down to the uniqueness of a limit, since we a limit of a sequence of functions at some particular $x \in \mathbb{R}$ is more or less just a regular ole' sequence.
