- #1

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If [tex]\lim_{n\to\infty}x_{n} = L[/tex] and [tex]x_{n}[/tex] is decreasing,then [tex]\forall n[n\in N\Longrightarrow x_{n}\geq L][/tex].

On the following proof give a list of all the thoerems ,axioms,definitions and the laws of logic that take place in the proof

proof:

Suppose that there exists [tex]n_1\in\mathbb N[/tex] such that [tex]x_{n_1}<L[/tex]. Take [tex]\epsilon=L-x_{n_1}[/tex]. Since [tex]\displaystyle\lim_{n\rightarrow\infty}x_n=L[/tex] there exists [tex]n_2\in\mathbb N[/tex] such that [tex]\|x_n-L\|<\epsilon[/tex] for all [tex]n\geq n_2[/tex]. Take [tex]n_0>\max\{n_1,n_2\}[/tex]. Then we have that [tex]x_{n_1}<x_n<2L-x_{n_1}[/tex] for all [tex]n\geq n_0[/tex] in particular, [tex]x_{n_1}<x_{n_0}[/tex]. But since [tex]n_1<n_0[/tex] and the sequence is decreasing we have that [tex]x_{n_0}<x_{n_1}[/tex] wich gives us a contradiction. Hence [tex]x_n\geq L[/tex] for all [tex]n\in\mathbb N[/tex].