# 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

•

Total views: 308.1k

•

Views today: 6.85k

Answer

Verified

308.1k+ views

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 .

Recently Updated Pages

If ab and c are unit vectors then left ab2 right+bc2+ca2 class 12 maths JEE_Main

A rod AB of length 4 units moves horizontally when class 11 maths JEE_Main

Evaluate the value of intlimits0pi cos 3xdx A 0 B 1 class 12 maths JEE_Main

Which of the following is correct 1 nleft S cup T right class 10 maths JEE_Main

What is the area of the triangle with vertices Aleft class 11 maths JEE_Main

KCN reacts readily to give a cyanide with A Ethyl alcohol class 12 chemistry JEE_Main

Trending doubts

What was the capital of Kanishka A Mathura B Purushapura class 7 social studies CBSE

Difference Between Plant Cell and Animal Cell

Write an application to the principal requesting five class 10 english CBSE

Ray optics is valid when characteristic dimensions class 12 physics CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Tropic of Cancer passes through how many states? Name them.

Write the 6 fundamental rights of India and explain in detail

Write a letter to the principal requesting him to grant class 10 english CBSE

Name the Largest and the Smallest Cell in the Human Body ?