Real Analysis: Proving a theorem for limit of functions help please

Question

I need to prove the equivalence relation: $\lim_{x \to c^+} f = L\Leftrightarrow $ for every sequence $(x_ n)$ that converges at $c$ s.t. $x_n\in A$ and $x_n>c$ $\forall n\in \mathbb{N}$, the sequence $(f(x_n))$ converges to $L$.

($\rightarrow$) Suppose $\lim_{x \to c^+} f = L$ and that $(x_ n)$ is a sequence in $A$ with $\lim (x_n) = c$ for $x_n>c$ $\forall n\in \mathbb{N}$. We must prove that $(f(x_n))$ converges to $L$.

Let $\epsilon>0$ be given. Then there exists a $\delta>0$ s.t. $\forall x\in A$ satisfying $0<x-c<\delta \Rightarrow$ $\vert{f(x)-L}\vert<\epsilon$.

Now, I have done the obvious part I guess. And now I have to somehow link the definition of sequence convergence to this(using $K(\delta)$ I think). How do I do this? Any help would be appreciated.

Answer 1

This statement is easier to prove by contrapositive. That is, show that if $\lim_{x \to c^+} f(x)$ either doesn't exist or is not equal to $L$, there must exist a sequence $x_n \to c$ (from the right) for which $\lim_{n \to \infty} f(x_n)$ either doesn't exist or is not equal to $L$.

In particular, if the limit from the right (does not exist or exists and) is not equal to $L$, we have the following statement:

There exists an $\epsilon>0$ such that for any $\delta > 0$, there is an $x$ such that $c < x < c + \delta$ and $|f(x) - L| \geq \epsilon$.

Try using this to construct the necessary sequence.