Question: Let ${(x_n)} $ be a sequence of real numbers and $x \neq 0$. $\lim_{n \to \infty} x_n =x$ iff: $\lim_{n \to \infty}(x_n-x)/(x+x_n)=0$



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: $\lim_{n \to \infty}(x_n-x)/(x+x_n)=0$

= ${\lim_{n \to \infty}(x_n-x)}/\lim_{n \to \infty}(x+x_n)=0/2x =0$


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 $$(x_n-x)/(x+x_n)=y_n$$


Assuming that $(x+x_x)$ is bounded,

$$lim _{n\to \infty} (x_n-x)= lim _{n\to \infty}(x+x_x)y_n =0$$

$$lim _{n\to \infty} (x_n-x)\implies lim _{n\to \infty} x_n=x$$

  • $\begingroup$ Also, why did you assume that $(x+x_n)$ is bounded? $\endgroup$ – A.Asad Mar 14 '18 at 10:16
  • $\begingroup$ Otherwise we have a case of $0\times \infty $ $\endgroup$ – Mohammad Riazi-Kermani Mar 14 '18 at 11:43
  • $\begingroup$ Hello, can you please have a look at this question of mine and verify if the proof is correct? Thanks in advance. $\endgroup$ – A.Asad Apr 11 '18 at 2:09
  • $\begingroup$ Yes, your forward implication is correct. $\endgroup$ – Mohammad Riazi-Kermani Apr 11 '18 at 2:26

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