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No. 77
>>73
First of all I don't know how to post properly...
As for "1. Prove that the limit of a sum is the sum of the limits. lim a + b = lim a + lim b for two sequences a and b, provided that both limits exist."
Let X be a subset of R, let E be a subset of X, let x[0] be an adherent point of E, and let f:X->R and g:X->R be functions. Now, suppose that f has a limit L at x[0] in E, and g has a limit M at x[0] in E. Then f+g has a limit L+M at x[0] in E since x[0] is an adherent point of E, then we can construct a sequence (a[n])n=0,infinity, consisting of elements in E, which converge to x[0]. Since f has a limit L at x[0] in E, (f(a[n]))n=0, infinity must converge to L. Similarly, (g(a[n]))n=0, infinity converges to M. By the limit laws for sequences (that is: lim of n->infinity (a[n]+b[n])= lim of n-> infinity a[n]+ lim of n->infinity b[n], since the sum of Cauchy sequences is Cauchy (Let x=lim n->infinty a[n] and y=lim n->infinity b[n] be real numbers. Then x+y is a lso a real number as for every epsilon>0 the sequence (a[n]+b[n])n=1, infinity is eventually epsilon-steady, show that (a[n]) is delta-steady, set delta equal to epsilon over 2, and show that this is eventually true for n>=N, similarly show this for (b[n]), this implies that a[n]+b[n] is eventually epsilon close when you add them) we conclude that ((f+g)(a[n]))n=0, infinity converges to L+M. this implies that f+g has a limit L+M at x[0] in E.
Sorry for getting a bit careless halfway, the typing was getting a little annoying, writing it down is much easier as you can clearly see everything.
PS: I just noticed I didn't directly answer your question, but also showed that this is true for added limits of functions :)
PPS: I just also realized I should answer in a picture... but last time I tried to upload a pic I got a temp. ban for some reason... sorry?
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