# Is this proof correct?

• Mar 10th 2010, 06:18 PM
tn11631
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 $x_0$ and L be real numbers, and let f be a real-valued function defined in a deleted neighborhood of $x_0$.

(a) What is the logical negation of $lim_x-> img.top {vertical-align:15%;} $x_0$" alt="lim_x-> $x_0$" />f(x)=L?

The def is: For all $\epsilon$>0 there exists $\delta$>0 for all x such that 0<|x- $x_0$|< $\delta$ $\Rightarrow$ |f(x)-L|< $\epsilon$

The negation is: There exists $\epsilon$>0 for all $\delta$>0 there exists x such that 0<|x- $x_0$< $\delta$ and |f(x)-L|>= $\epsilon$

b is the one I need checked:
(b) Show that lim_x-> $x_0$f(x)=L if and only if for every sequence { $x_n$}n=1 to infinity with lim_x-> $x_n$= $x_0$ with $x_n$ $\neq$ $x_0$, it is true that lim_x->infinity f( $x_n$=L

Here is my proof:
Suppose $lim_x-> img.top {vertical-align:15%;} $x_0$" alt="lim_x-> $x_0$" />f(x)=L. Let { $x_n$}n=1 to infinity be a sequence where $x_n$ $\neq$ $x_0$ but lim_n-> infinity $x_n$= $x_0$ and consider the sequence {f( $x_n$)}n=1 to infinity . choose $\epsilon$>0/ there is delta>0 such that if 0<|x- $x_0$|<delta then |f(x)-L|< $\epsilon$. Since {x_n} n=1 to infinity converges to $x_0$ there is N such that for n>=N, | $x_n$- $x_0$|<delta. Now for n>=N 0<| $x_n$- $x_0$|<delta. Hence, |f( $x_n$)-l|< $\epsilon$. 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 $x_0$. Thus, there is $\epsilon$>0 such that for every delta>0, 0<|x-x_0|<delta and such that |f(x)-l|>= $\epsilon$. In particular for each positive integer n, 0<| $x_n$- $x_0$)<1/n such that |f(x_n)-L|>= $\epsilon$. The sequence { $x_n$}n=1 to infinity converges to $x_0$ and is a sequence where $x_n$ is NOT = $x_0$, hence {f(x_n)}n=1 to infinity converges to L which contradicts the fact that |f(x_n)-L|>= $\epsilon$>0 for all n. Thus L must be the limit of f at $x_0$.
• Mar 11th 2010, 12:13 PM
Tinyboss
Looks good, but work on your LaTeX! ;-)