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For what values of $ x $ is $ f $ continuous?

$$ f(x) = \left\{

\begin{array}{ll}

0 & \mbox{if $ x $ is rational}\\

1 & \mbox{if $ x $ is irrational}

\end{array} \right.$$

$f(x)=\left\{\begin{array}{l}0 \text { if } x \text { is rational } \\ 1 \text { if } x \text { is irrational }\end{array} \text { is continuous nowhere. For, given any number } a \text { and any } \delta>0 \text { , the interval }(a-\delta, a+\delta)\right.$

contains both infinitely many rational and infinitely many irrational numbers. since $f(a)=0$ or 1 , there are infinitely many

numbers $x$ with $0<|x-a|<\delta$ and $|f(x)-f(a)|=1 .$ Thus, $\lim _{x \rightarrow a} f(x) \neq f(a) .$ [ In fact, $\lim _{x \rightarrow a} f(x)$ does not even exist.

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This is problem number sixty seven of this tour Calculus eighth edition, Section two point five for what values of X is f continuous half of X is equal to zero. Effects is rational. And one, if x is irrational. And we're going to discuss which values of X are continuous on F as well as which ones are discontinuous and, ah, it may help us to. I have a visual interpretation over this function might look like, um, for any rational value. Let's choose, for example, Um, okay, we'LL say rational number of three. The ah, this X would have a value of zero for the function and not to ah, faraway from three is an irrational number. Kanpai, or the values of function according to athletics. As it is, ah, to find would have a value one. So this is an example of what we might see the function behaving. It is either zero if Thie values of X are rational, for example, like one, two three and irrational. If it's ah, Iraq and and if it's irrational, the function is value one such as pie and maybe squirt of two. Andi already see that the behavior seems discontinuous, but we will go ahead and discuss that a little further. The definition of continuity for a function f is that the limit and sex approaches a of a function and must be equal to every day. So we say that the function of his continuous if and only if this is the case and the main main issue that we run into is actually with not with confirming that the Limited is equal to the function evaluated a but rather whether this funk with whether this limit is to find her Now, Okay, this limit we can take at any point. Let's choose three. For example, if we were to determine, try to determine what the limit is for point three. We we need to do two things. We need to find the limit as X approaches three from the left trying to limit as ex purchased three from the right. And not only those two. Those two values have tio exist, but they have to be equal to each other and own. And the limit above is equal is equal to the limits of thes air equal Teo If and only if these left and right limits exist, and they are equal to each other. But what we can see here is there. When we look at X opportunity approaching three, which is a rational number, this is a rational example. It's uncertain whether from the left or the right, we are approaching from another rational number or an irrational number, and that is actually known that in the near vicinity of irrational number, there are infinitely many irrational numbers. Try being one example, but there are in infinite, many of rush irrational numbers that we are not. We don't have the time to to find at the moment, but we know that there are irrational numbers around three immediately in the vicinity of three, such that their values will be equal to one because they are arrested. And if that's the case, then we see that the limit from the left and the right does not exist because of that disk continuity. So the limit from left is that exist in that one right does not exist. They can't equal each other, essentially the limit. The limit is ex Purchase three does not exist, and as such, we can confirm continuity at that exact point for three and this is the same for any other rational number. If we do this exact same no example for an irrational number such as pie, we use the same logic because we understand that it is given that near the vicinity of any irrational number such as pirates. Hard too. There are infinitely many amount of rational members for a pie. For example, it is near the vicinity of three point one and three point two or three point one, three hundred point one five, which are rational numbers. And so those values of the function will be equal to zero, whereas at pi the value of the functions one. So this discontinuity leads to the fact that the limit does not exist, and since the limit is not exist, it can't be equal to the function evaluated at the value of a meaning that you cannot confirm continuity. And so whether or not we approach this using national numbers or irrational numbers dysfunction in general ah is definite. It definitively not continuous at any value of X