Limit of a sequence ${a_n}$

Question

Let {${a_n}$} be a sequence of real numbers and $a \ne 0$.

$\lim_{n \to \infty} {a_n} = a$ iff $\lim_{n \to \infty} (\frac{a-a_n}{a+a_n}) = 0$

My approach:

Forward direction:

$\lim_{n \to \infty} (\frac{a-a_n}{a+a_n})$

$= (\frac{a-a}{a+a}) $

$= 0$

I don't know how to show the implication in the other direction. Also is my approach for the forward direction correct?

Answer 1

Hint:   if $\,\frac{a-a_n}{a+a_n} = \frac{2a}{a+a_n}-1 \to 0\,$ then $\,\frac{2a}{a+a_n} \to 1\,$ so $\,\frac{a+a_n}{2a} \to 1\,$ and $\;\ldots$

Answer 2

Let $b_n=\frac {a-a_n}{a+a_n}.$ Then $a_n=\frac {a(1-b_n)}{1+b_n}.$ If $b_n\to 0$ then by the forward direction we have $$a=a\cdot 1=a\cdot \lim_n\frac {1-b_n}{1+b_n}=\lim_n\frac {a(1-b_n)}{1+b_n}=\lim_n a_n.$$