proving pointwise convergence formally I'm having trouble formally proving this simple sequence of functions is pointwise continuous, I know the limit should be $0$.
$f_n(x) = 0$ for $ x < n$ and $f_n(x) = 1/x$ for $ x \geq n$ Could someone show me how to formally prove it?
 A: Take any point $x$ under consideration. You have (correctly) guessed that the limit is $f \equiv 0$. Thus, you need to demonstrate, that for every $\epsilon > 0$, there is $N$ so that 
$$
|f_n(x) - f(x)| = |f_n(x)| \leq \epsilon
$$
if $n \geq N$. 
In particular, what happens if you choose $N > x$?
(further, you can prove this sequence of functions is uniformly continuous, meaning the $N$ doesn't need to depend on $x$: for every $\epsilon > 0$, you can find $N$ so that $|f_n(x)| < \epsilon$ for all $x$ and $n \geq N$)
To elaborate: To demonstrate pointwise convergence, choose $x$ first. Then choose $N$. In particular, if you choose $N > x$, then $n > x$ as well, and you never have to consider $x \geq n$.
If you instead want to demonstrate uniform convergence, choose your $\epsilon > 0$, then choose $N$ so that $N > 1/\epsilon$. Now, choose any $x$ and any $n \geq N$. We have to show that $|f_n(x)| < \epsilon$. 
So, either $x < n$ or $x \geq n$. If $x < n$, then $f_n(x) = 0$ and you are done. Otherwise, $$|f_n(x)| = \dfrac1x \leq \dfrac1n \leq \dfrac1N < \epsilon.$$
What is the difference here?
For pointwise convergence, you can choose $N$ that depends on $x$ and $\epsilon$.
For uniform convergence, you must choose $N$ that only depends on $\epsilon$ and not on $x$.
