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# Let $f:[0,1]\to R$ be defined by f(x)=\left\{ \begin{align} & 1,\text{ if }x\text{ is rational} \\ & 0,\text{ if }x\text{ is irrational} \\ \end{align} \right.Show that the limit does not exist.

Last updated date: 30th Mar 2023
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Hint: For every number $x$ that is rational, there exists a value in the neighborhood of $x$, i.e. $(x-\Delta ,x+\Delta )$ which is irrational. Similarly, for every number $y$ that is irrational there exists a value in the neighborhood of $y$, i.e. $(y-\Delta ,y+\Delta )$ that is rational.

The given function is f(x)=\left\{ \begin{align} & 1,\text{ if }x\text{ is rational} \\ & 0,\text{ if }x\text{ is irrational} \\ \end{align} \right. . We need to show that the limit does not exist for the given function .
Let us assume that the limit exists. So , in such case , there should be a number $\Delta >0$, such that for all $y\in R$ with $\left| y-x \right|<\Delta$ , it should satisfy the condition $\left| f(y)-f(x) \right|<1$.
i.e. $\forall y\in (x-\Delta ,x+\Delta ),\left| f(y)-f(x) \right|<1$
Now , if $x$ is rational , there exists $y\in (x-\Delta ,x+\Delta )$ such that $y$ is irrational. For example , $1.4=\dfrac{14}{10}$ is a rational number and in its neighborhood , we have $\sqrt{2}=1.4142...$ which is irrational .
So, $\left| f(y)-f(x) \right|=\left| 0-1 \right|=1$
i.e. $\left| f(y)-f(x) \right|{<}1$
Hence , we can clearly see that it contradicts the statement that if the limit exists then there should be a number $\Delta >0$, such that for all $y\in R$ with $\left| y-x \right|<\Delta$ , it should satisfy the condition $\left| f(y)-f(x) \right|<1$ .
Hence , the statement that the limit exists is wrong , or we can simply say that the limit does not exist .

Note: If $x$ is irrational, there exists $y\in (x-\Delta ,x+\Delta )$ which is rational . For example , $\sqrt{2}=1.4142...$ is an irrational number and in its neighborhood , we have $1.4=\dfrac{14}{10}$which is rational . So , $\left| f(y)-f(x) \right|=\left| 0-1 \right|=1$ , which , again , contradicts the fact $\left| f(y)-f(x) \right|<1$. So , it will contradict the statement that if the limit exists then there should be a number $\Delta >0$, such that for all $y\in R$ with $\left| y-x \right|<\Delta$ , it should satisfy the condition $\left| f(y)-f(x) \right|<1$ . Hence , the limit does not exist .