@Matt I.e., for any $x$, for any $\varepsilon$, there exists $N$ such that the world is nice after that point. The point here is that the $N$ depends on $x$.
Let's look at what your post says. Ignore the first bit, it's not critical. In the second case, you argue that $\cos x$ being less than $1$, $(\cos x)^n \to 0$. True, but this is a pointwise statement.
We are interested in how large $n$ should be so that $f_n(x) < \varepsilon$, yes?
@Matt Um, you're right -- in a sense. The issue is not that we cannot speak of uniform convergence when the limit function isn't continuous.
The more precise way of saying what you said is this: Theorem: if $f_n(x)$ is continuous over $[0,1]$ and $f_n(x)$ converges to $f(x)$ uniformly, then $f(x)$ is continuous.
That is a correct explanation, but it's too clever for my purpose here.
@Matt True. But the theorem's converse is not necessarily true, so you cannot apply it in the reverse to conclude that the $f_n(x)$ in that post has to converge uniformly.
@Srivatsan Let me try again: for a fixed $\varepsilon$ you can find $x$ and $y$ such that $|f(x) - f(y)| > \varepsilon$. Namely, for $x = 1$ and any other point $y$ in $[0,1]$. Because as $n$ gets large, $y^n$ gets arbitrarily small?
@Srivatsan I can't think of how to make it more precise but I'll try to rephrase it: Fix $\varepsilon > 0$ and pick $x=1$ and any $y \in [0,1)$. Then $|f(x) - f(y)| = |1 - y^n|$. The limit of this is $1$ so no matter what $N$ you pick, this is never smaller than $\varepsilon$ so $f_n$ doesn't converge uniformly.
:) HINT: For a fixed $\lambda$, what does the sequence $(1 + \frac{\lambda}{n})^n$ converge to?
@Matt Alternatively, let's look at the multiplicative inverse of the sequence: $$ \left( \frac{n}{n-1} \right)^n = \left( 1 + \frac{1}{n-1} \right)^n .$$ What does that converge to?
If anything is not clear, I will explain more clearly.
