Show that if limit of sequence $x_n$ goes to $x$ as $n$ goes to $ \infty$ and $ x > a$, then $x_n > a$ all but finitely many $ n$.

I have started it like this: $x_n > x - \epsilon$ (as $|x_n- x| < \epsilon$ for all $\epsilon$)

$x_n > a - \epsilon$

From here on, how does one chose the right epsilon to show that $x_n > a$?


It is always best to draw a number line (if possible do it mentally rather than on paper) and position $x, a$ on it: $$------a-----x-----$$ Now things are crystal clear. As $n\to\infty$ the values of $x_n$ will come near $x$ and from the figure above you can see that as one moves near $x$ one gets greater than $a$.

The question is "how near should one go to $x$ in order to become greater than $a$?". Since the distance between $x$ and $a$ is $|x-a|=x-a$ one has to move closer to $x$ by anything less than this amount. Choose any positive $\epsilon$ less than $x-a$ and you are done. In particular you can choose $\epsilon=(x-a) /2$. Note that $\epsilon=x-a$ also works because the inequality used in limit definition is $<$ and not $\leq $.

  • $\begingroup$ Great answer. (+1) $\endgroup$ – Math_QED Sep 20 '18 at 7:31
  • $\begingroup$ @Math_QED: glad to know someone liked it very much. In general most of the $\epsilon, \delta$ stuff in introductory analysis is easy once you think of inequalities on the number line. Often people do the opposite and think in terms of symbol manipulation and perhaps that's one of the reasons most beginners find such proofs difficult. $\endgroup$ – Paramanand Singh Sep 20 '18 at 9:28
  • $\begingroup$ Thank you very much! $\endgroup$ – Aishwarya Deore Sep 20 '18 at 10:39
  • $\begingroup$ @AishwaryaDeore: in case you like any of the answers here, you may accept it. $\endgroup$ – Paramanand Singh Sep 20 '18 at 11:40

Let $\epsilon = \frac {x-a}{2}$

Then $x-\epsilon = \frac {x+a}{2} >a$

Thus $x_n > x-\epsilon \implies x_n >a$

  • $\begingroup$ Thank you very much! $\endgroup$ – Aishwarya Deore Sep 20 '18 at 10:39

If a sequence $x_n$ converges to $x$, by definition it means for every $\epsilon >0$ there exist a $N$ such that $$|x_n-x|<\epsilon\tag{1}$$ for all $n>N$

Now (1) can be rephrased as $x_n\in (x-\epsilon, x+\epsilon)\tag{2}$ for all $n>N$

Take $\epsilon=x-a$ (as $x>a, x-a$ is a positive number and hence the choice of taking $x-a=\epsilon$ makes sense).

By definition there exist a $N$ such that $$x_n\in (x-\epsilon, x+\epsilon)$$ for all $n>N$. Now put the value of $\epsilon $ you get $$x_n\in (x-(x-a),x+x-a)=(a,2x-a)$$

Therefore $x_n>a$ for all $n>N$

  • 1
    $\begingroup$ Why the downvote? +1 to compensate. $\endgroup$ – Paramanand Singh Sep 20 '18 at 2:53
  • $\begingroup$ Well the upvote was also to indicate that your answer is correct apart from the compensation. $\endgroup$ – Paramanand Singh Sep 20 '18 at 3:05
  • $\begingroup$ Thank you very much! $\endgroup$ – Aishwarya Deore Sep 20 '18 at 10:40

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