Answer
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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 .
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 .