
Is this proof correct?
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 realvalued function defined in a deleted neighborhood of .
(a) What is the logical negation of
img.top {verticalalign: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 {verticalalign: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<xx_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 .

Looks good, but work on your LaTeX! ;)