Question: Let ${(x_n)} $ be a sequence of real numbers and $x \neq 0$. $\lim_{n \to \infty} x_n =x$ iff: $\lim_{n \to \infty}(x_n-x)/(x+x_n)=0$
Proof:
"$\implies$"
Since $\lim_{n \to \infty} x_n =x$. This implies that $\forall \epsilon>0$, $\exists N \in \mathbb{N}$ such that"
$|x_n - x|<\epsilon$ $\forall n\geq N$
$ \implies |(x_n -x) - 0|<\epsilon$ $ \forall n\geq N$
$\implies$ $\lim_{n \to \infty} (x_n -x)=0$
Now consider $|x +x_n|$
Claim: $\lim_{n \to \infty} |x+x_n| =2x$
Let $\epsilon>0$ be given
$|(x+x_n)-2x| = |x_n -x| < \epsilon$ $\forall n\geq N$ for some$N \in \mathbb{N}$
Then: $\lim_{n \to \infty}(x_n-x)/(x+x_n)=0$
= ${\lim_{n \to \infty}(x_n-x)}/\lim_{n \to \infty}(x+x_n)=0/2x =0$
"$\Longleftarrow$"
Can anyone please verify the forward implication of the proof? Also, I'm unable to see how to go about the backward implication. Can anyone please guide/give hint?
Thank you.
Your forward implication is correct.
$$\lim_{n \to \infty}(x_n-x)/(x+x_n)=0$$
Let $$(x_n-x)/(x+x_n)=y_n$$
$$(x_n-x)=(x+x_x)y_n$$
Assuming that $(x+x_x)$ is bounded,
$$lim _{n\to \infty} (x_n-x)= lim _{n\to \infty}(x+x_x)y_n =0$$
$$lim _{n\to \infty} (x_n-x)\implies lim _{n\to \infty} x_n=x$$