Consider the function $f$:R$\rightarrow $R defined by $$f(x) =\begin{cases}x-1, &\text{if $x$ is rational} \\5-x,&\text{if $x$ is irrational}\end{cases}$$ Then $\space\lim\limits_{x\to a}$$f(x)$, $a\in\ R-\{\ 3\}$, exists or not ?

Solution: Let $a$ be a irrational number .Then

Right hand limit and left hand limit are as follows;

$\space\lim\limits_{x\to a^+}$$f(x)$ =$\space\lim\limits_{h\to 0}$$f(a+h)$; $\space$$\space\lim\limits_{x\to a^-}$$f(x)$ =$\space\lim\limits_{h\to 0}$$f(a-h)$

As h$\rightarrow$$0$, now let us assume that $h$ be a rational number, then $a+h$ and $a-h$ both are irrational . Therefore

R.H.L. =$\space\lim\limits_{x\to a^+}$$f(x)$ =$\space\lim\limits_{h\to 0}$$f(a+h)$=$\space\lim\limits_{h\to o}$$5-(a+h)$$\space$=$\space$$5-a$



Hence the limit exists.

Now again let us assume that $h$ be a irrational then $a+h$ and $a-h$ may be a rational or irrational, then the L.H.L. and R.H.L. may or may not be equal and hence limit may or may not be exist.But in my booklet the question says that the limit exists only if $a=3$. Is it true or wrong ?

  • $\begingroup$ Hi and welcome to the site! Since this is a site that encourages and helps with learning, it is best if you show your own ideas and efforts in solving the question. Can you edit your question to add your thoughts and ideas about it? $\endgroup$ – 5xum Jun 2 '16 at 8:24
  • $\begingroup$ You should provide more details. For example, what have you tried on this so far? What's your opinion etc. etc... $\endgroup$ – BigbearZzz Jun 2 '16 at 8:24
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    $\begingroup$ Also, don't get discouraged by the downvote. I downvoted the question and voted to close it because at the moment, it is not up to site standards (you have shown no work you did on your own). If you edit your question so that you show what you tried and how far you got, I will not only remove the downvote, I will add the upvote. $\endgroup$ – 5xum Jun 2 '16 at 8:24
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    $\begingroup$ Ok, I am sharing my solution and then I will ask my doubt. $\endgroup$ – Girish Kumar Chandora Jun 2 '16 at 8:29
  • $\begingroup$ anyone please tell me am I right or wrong $\endgroup$ – Girish Kumar Chandora Jun 2 '16 at 10:35

The function converges to $a-1$ on rationals, while it converges to $5-a$ on irrationals. So the function converges on the reals iff the two limits coincide, $a-1=5-a$, i.e. $a=3$.


Let $a\in\mathbb R$. Then there exists a sequence $q_n = \frac{\lfloor n a+1\rfloor}n$

Then $q_n\neq a\ \forall\ n\in N$. Also $q_n\in\mathbb Q$ (rational) for all $n$.

As we know that

$$x-1<\lfloor x\rfloor\leq x\qquad \forall\ x\in\mathbb R.$$

From this it follows

\begin{align*} \frac{na+1-1}{n}&<\frac{\lfloor na+1\rfloor}{n}\leq \frac {na+1}{n}\\ \lim_{n\to\infty}\frac{na+1-1-1}{n}&\leq\lim_{n\to\infty}\frac{\lfloor na+1\rfloor}{n}\leq\lim_{n\to\infty} \frac {na+1}{n}\\ a&\leq \lim_{n\to\infty}\frac{\lfloor na+1\rfloor}{n}\leq a \end{align*}

and by applying the sandwich/squeeze theorem we get


Similary there exists the sequence $r_n = q_n + \frac{\sqrt{2}}n$

We have $$r_n\neq a\qquad \forall\ n\in \mathbb N.$$

Further $r_n\to a$ for $n\to\infty$ and $r_n\in\mathbb R\setminus\mathbb Q$ (irrational) for all $n$. Now we look at $$ \lim_{n\to\infty}f(q_n) = \lim_{n\to\infty} q_n - 1 = a-1 $$ and $$ \lim_{n\to\infty}f(r_n) = \lim_{n\to\infty} 5 - r_n = 5-a $$ If the limit $\lim_{x\to a}f(x)$ should exist, both of the above limites should give the same value. This is obviously not the case for $a\neq 3$.

  • $\begingroup$ partially satisfied with your solution, you assumed the sequence q' whose term may be equal to the a. but according to the concept the term must not be equal to the a, for any value of N. Further, you also assumed that a is a real number then it may be rational or irrational. how can you say that the term of the sequence q are rational numbers..... I can't understand.. further I am sharing my alternative solution $\endgroup$ – Girish Kumar Chandora Jun 3 '16 at 8:41
  • $\begingroup$ Yes $a$ may be irrational or rational. $q_n$ is always rational, because the floor function $\lfloor\bullet\rfloor$ always gives back an integer and integer divided by integer is always a rational number. Actually we can take any rational sequence $q_n\to a$ and any irrational sequence $r_n\to a$, the argumentation would still hold. I just gave examples for them in order to show that there exist sequences of both kinds. $\endgroup$ – Wauzl Jun 3 '16 at 8:51
  • $\begingroup$ i apologies for it that i did not seen the floor function properly and saw it as absolute function $\endgroup$ – Girish Kumar Chandora Jun 3 '16 at 8:55
  • $\begingroup$ Also, the sequences may contain $a$ as an element. Even the constant sequence $x_n = a$ is valid (but in this context it doesn't help to use this particular one, as you don't know whether $a$ is rational or not) $\endgroup$ – Wauzl Jun 3 '16 at 8:55
  • $\begingroup$ you are saying that the sequences may contain a as an element. but according to the concept the term must not be equal to the a for any value of N. I have still confusion $\endgroup$ – Girish Kumar Chandora Jun 3 '16 at 9:06

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