Thread: Is this proof correct?

1. 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 $\displaystyle x_0$ and L be real numbers, and let f be a real-valued function defined in a deleted neighborhood of $\displaystyle x_0$.

(a) What is the logical negation of $\displaystyle lim_x->$\displaystyle x_0$$f(x)=L? The def is: For all \displaystyle \epsilon>0 there exists \displaystyle \delta>0 for all x such that 0<|x-\displaystyle x_0|<\displaystyle \delta \displaystyle \Rightarrow |f(x)-L|<\displaystyle \epsilon The negation is: There exists \displaystyle \epsilon>0 for all \displaystyle \delta>0 there exists x such that 0<|x-\displaystyle x_0<\displaystyle \delta and |f(x)-L|>= \displaystyle \epsilon b is the one I need checked: (b) Show that lim_x->\displaystyle x_0f(x)=L if and only if for every sequence {\displaystyle x_n}n=1 to infinity with lim_x->\displaystyle x_n=\displaystyle x_0 with \displaystyle x_n$$\displaystyle \neq$$\displaystyle x_0, it is true that lim_x->infinity f(\displaystyle x_n=L Here is my proof: Suppose \displaystyle lim_x->\displaystyle x_0$$f(x)=L. Let {$\displaystyle x_n$}n=1 to infinity be a sequence where $\displaystyle x_n$$\displaystyle \neq$$\displaystyle x_0$ but lim_n-> infinity $\displaystyle x_n$=$\displaystyle x_0$ and consider the sequence {f($\displaystyle x_n$)}n=1 to infinity . choose $\displaystyle \epsilon$>0/ there is delta>0 such that if 0<|x-$\displaystyle x_0$|<delta then |f(x)-L|<$\displaystyle \epsilon$. Since {x_n} n=1 to infinity converges to $\displaystyle x_0$ there is N such that for n>=N, |$\displaystyle x_n$-$\displaystyle x_0$|<delta. Now for n>=N 0<|$\displaystyle x_n$-$\displaystyle x_0$|<delta. Hence, |f($\displaystyle x_n$)-l|<$\displaystyle \epsilon$. 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 $\displaystyle x_0$. Thus, there is $\displaystyle \epsilon$>0 such that for every delta>0, 0<|x-x_0|<delta and such that |f(x)-l|>= $\displaystyle \epsilon$. In particular for each positive integer n, 0<|$\displaystyle x_n$-$\displaystyle x_0$)<1/n such that |f(x_n)-L|>= $\displaystyle \epsilon$. The sequence {$\displaystyle x_n$}n=1 to infinity converges to $\displaystyle x_0$ and is a sequence where $\displaystyle x_n$ is NOT = $\displaystyle x_0$, hence {f(x_n)}n=1 to infinity converges to L which contradicts the fact that |f(x_n)-L|>=$\displaystyle \epsilon$>0 for all n. Thus L must be the limit of f at $\displaystyle x_0$.

2. Looks good, but work on your LaTeX! ;-)