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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$

Similiarly

L.H.L.$\space$=$5-a$

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 ?

JMP
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Girish Kumar Chandora
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  • 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? – 5xum Jun 02 '16 at 08:24
  • You should provide more details. For example, what have you tried on this so far? What's your opinion etc. etc... – BigbearZzz Jun 02 '16 at 08:24
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    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. – 5xum Jun 02 '16 at 08:24
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    Ok, I am sharing my solution and then I will ask my doubt. – Girish Kumar Chandora Jun 02 '16 at 08:29
  • anyone please tell me am I right or wrong – Girish Kumar Chandora Jun 02 '16 at 10:35
  • Hint: If you know that $a$ is irrational, you don't know if $a+h$ is rational or irrational, so you can't use the irrational formula for $f$ on $a+h$. – Michael Burr Jun 02 '16 at 12:35

2 Answers2

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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

$$\lim_{n\to\infty}q_n=a.$$

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 $a\neq 3$.

Stefan Hante
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  • 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 – Girish Kumar Chandora Jun 03 '16 at 08:41
  • 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. – Stefan Hante Jun 03 '16 at 08:51
  • i apologies for it that i did not seen the floor function properly and saw it as absolute function – Girish Kumar Chandora Jun 03 '16 at 08:55
  • 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) – Stefan Hante Jun 03 '16 at 08:55
  • 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 – Girish Kumar Chandora Jun 03 '16 at 09:06
  • now you can see that no term of the sequence is equl to the $a$. i have edited your solution.... – Girish Kumar Chandora Jun 03 '16 at 11:50
  • You did a nice job showing that $q_n\to a$ actually holds. Still I'm saying that for showing that $\lim_{x\to a} f(a) = Y$ exists, for all sequences $x_n\to x$ including those who have $a$ as elements it has to hold $\lim_{n\to\infty} f(x_n) = Y$. Thus it is not necessary to prove that $q_n\neq a$. – Stefan Hante Jun 03 '16 at 12:20
  • but if we restrict the domain of f to $R-{a}$, then your assumed sequence not holds good why because it contains a and according to the theory for the assumption of sequence for $f:S\rightarrow R$ must be S. Here set S=R, so I also agree to your assumed sequence – Girish Kumar Chandora Jun 04 '16 at 08:23
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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. $a=3$.