Verified answer

I'm a bit confused by the notation, but if I'm understanding it correctly then, seeking a contradiction, suppose that as n → ∞ then (n, 1/n) → (L₁, L₂) which by a result in analysis (it is really easy to prove) is equivalent to saying that as:

(1) n → ∞ n → L₁

(2) n → ∞ 1/n → L₂

(i.e. (x, y) → (x', y') if and only if x → x' and y → y')

The first is clearly false because as n tends to infinity n doesn't converge. The second does converge (namely to zero), but since the first doesn't converge then P_n doesn't converge. If you need to prove each of those rigorously then it isn't very hard to do.

Hope this helps!

p.s. What I'm confused is that is this a coordinate sequence? Or is the sequence you're considering simply 1/n?

it truly is effect of appropriate the following theorem enable f_n be a set of purposes defined on a subset A of R^n that converges uniformly to a function f with selection in R^m. enable a be a reduce element of A such that, for each n, L_n = lim (x ? a) f_n(x) exists in R^m. Then, f has a reduce at a and lim (x ? a) f(x) = lim L_n (actually, this theorem holds in favourite metric areas). it truly is, lower than uniform convergence, we are able to interchange the order the bounds are taken: lim (n ? ?) lim (x ? a) f_n(x) = lim (x ? a) lim (n ? ?) f_n(x) Polynomials are continuous on R and, hence, have a reduce at each and every authentic volume (merely the cost of the polynomial at this volume). quite at 0, this reduce is the self sufficient time period. considering 0 is a reduce element of (0,a million) it follows that if a set of polynomials converge uniformly on (0,a million) to a function f, then f has a reduce at 0+-. yet f(x) = sin(a million/x) shouldn't have a reduce at 0+, it keeps oscillating from -a million to at least a million and would not attitude any authentic volume. So, it won't be able to be the uniform reduce of a set of polynomials defined on (0, a million).