I can't find any similar questions online (except it might just be a bit too wordy), but say you are given a function:
$f_n(x) = \frac{1}{1+x^n}, x \in \mathbb{R}$
And the limit of $f_n(x)$ does not exist for x = -1 (and $n$ is odd), is this a sufficient reason for it to not uniformly converge (or even pointwise converge)?
I'm presuming the answer is yes but would just like a confirmation. It makes sense since the definition of uniform convergence is:
Given $\epsilon > 0 \exists N \in \mathbb{N}$ s.t. $\forall n>n. \left| f_n(x)-f(x) \right| < \epsilon$
But this is impossible if $f(x)$ is undefined.
So in short: if there is an $x$ value for which $f_n(x) \rightarrow f$ is undefined, then $f_n(x)$ cannot (uniformly) converge.
yes, it doesn't converge pointwise at -1, and subsequently the sequence doesn't converge uniformly on the set that includes -1.
check your quantifiers, too, in your definition of uniform convergence.
you should have, right after epsilon positive, "for all x in the set ... "
this asserts that the convergence is independent of the choice of x.