if-the-function-f-is-defined-by-fx-left-beginarrayll-0-mboxif-x-is-rational-1-mboxif-x-is-irrational

Question

If the function $ f $ is defined by
$$ f(x) = \left\{
          \begin{array}{ll}
              0  & \mbox{if $ x $ is rational}\
              1  & \mbox{if $ x $ is irrational}
           \end{array} ight.$$

prove that $ \displaystyle \lim_{x 	o 0}f(x) $ does not exist.

Daniel Jaimes · Accepted Answer

This is problem thirty nine of this to Rick Calculus Safe Edition section two point four. If the function F is defined by F of X is equal to zero, if excessive is rational, one of X is irrational. Prove that the limit is experts. Zero of the function f does not exist. We recall the definition of the limit on using Del Delta and Absalon, and here&#x27;s a graphical representation of that definition. We see that poor given function as you approach X. You should approach the limit. L, um provided that there are there exists now use for Delta For every given. Absalon, for example, here&#x27;s a closer minus epsilon tolerance for L, and we see that it corresponds to a certain TelDta tolerance for X. Now, as absolute get smaller, we should see that it also has a corresponding, smaller delta range. And so in this case, we would be able to prove that the limit for this function exists from because of this consistent epsilon. Delta found it parameters they were able to find. If we deal with this function, half let&#x27;s plot a little bit of what this might look like. Affects zero If X is rational. So let&#x27;s pick some rational values. How about well, at one over here it&#x27;s rational, the value one half his rational. Maybe the value in making it a zero point one. Jealous, irrational. And, of course, Ciro. I would also be rational. And because these are rational points, the function is given a value zero breach of these points. Irrational numbers are a very large set of numbers, and in fact, there are an infinite amount of irrational numbers between any two rational numbers. For example, anywhere around here wait, what? I need to know specifically what that irrational number is, but we know that there will be some points where in between a couple of rational numbers, there is an issue irrational number, meaning that the function we&#x27;ll have a value of one. And so what we see is dysfunction is definitely not continuous. It has values of between their own one, and it jumps very often. So it&#x27;s very Unser. And as we get closer to zero, this is where we&#x27;re trying to prove whether this exists. We see that we cannot define a certain Absalon tolerance for any assault tolerance we can not to find a certain Delta tolerance, which is what our definition of a limit is currently, Um, as we let&#x27;s say, we set a certain Delta tolerance around here. We see that this I could not allow for inappropriate, absolutely tolerance in order for us to define this limit. Andi element in this case does not exist because of the nature of the function as it jumps around a lot because there are an infinite many of international numbers, as well as rational members to the left, into the right of zero. And for this, for these explanations in these reasonings, we say that the limit, his expertise here of dysfunction does not exist.

Problem

By comparing Definitions 2, 3, and 4, prove Theor…

Problem 39 Medium Difficulty

If the function $ f $ is defined by
$$ f(x) = \left\{
\begin{array}{ll}
0 & \mbox{if $ x $ is rational}\\
1 & \mbox{if $ x $ is irrational}
\end{array} \right.$$

prove that $ \displaystyle \lim_{x \to 0}f(x) $ does not exist.

Answer

$$\text { This contradicts } L<\frac{1}{2}, \operatorname{so} \lim _{x \rightarrow 0} f(x) \text { does not exist}$$

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