Looks good, but work on your LaTeX! ;-)
Hey guys I just need some one to check this proof. Its sort of my weakness so if someone could just double check it. and please forgive the minor symbol mistakes, I tried
There are two parts to this problem but I got the first one with no problem but for the sake of the question I'm going to post everything.
Let and L be real numbers, and let f be a real-valued function defined in a deleted neighborhood of .
(a) What is the logical negation of img.top {vertical-align:15%;} " alt="lim_x-> " />f(x)=L?
The def is: For all >0 there exists >0 for all x such that 0<|x- |< |f(x)-L|<
The negation is: There exists >0 for all >0 there exists x such that 0<|x- < and |f(x)-L|>=
b is the one I need checked:
(b) Show that lim_x-> f(x)=L if and only if for every sequence { }n=1 to infinity with lim_x-> = with , it is true that lim_x->infinity f( =L
Here is my proof:
Suppose img.top {vertical-align:15%;} " alt="lim_x-> " />f(x)=L. Let { }n=1 to infinity be a sequence where but lim_n-> infinity = and consider the sequence {f( )}n=1 to infinity . choose >0/ there is delta>0 such that if 0<|x- |<delta then |f(x)-L|< . Since {x_n} n=1 to infinity converges to there is N such that for n>=N, | - |<delta. Now for n>=N 0<| - |<delta. Hence, |f( )-l|< . Thus, {f(x_n)}n=1 to infinity converges to L.
Suppose now that the second condition is satisfied and so all the sequences {f(x_n)}n=1 to infinity have a common limit called L. Suppose that L is NOT a limit of f at . Thus, there is >0 such that for every delta>0, 0<|x-x_0|<delta and such that |f(x)-l|>= . In particular for each positive integer n, 0<| - )<1/n such that |f(x_n)-L|>= . The sequence { }n=1 to infinity converges to and is a sequence where is NOT = , hence {f(x_n)}n=1 to infinity converges to L which contradicts the fact that |f(x_n)-L|>= >0 for all n. Thus L must be the limit of f at .