# 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.

Answer

Verified

382.2k+ 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

Basicity of sulphurous acid and sulphuric acid are

Why should electric field lines never cross each other class 12 physics CBSE

An electrostatic field line is a continuous curve That class 12 physics CBSE

What are the measures one has to take to prevent contracting class 12 biology CBSE

Suggest some methods to assist infertile couples to class 12 biology CBSE

Amniocentesis for sex determination is banned in our class 12 biology CBSE

Trending doubts

Difference Between Plant Cell and Animal Cell

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

What is the past tense of read class 10 english CBSE

Change the following sentences into negative and interrogative class 10 english CBSE

What is pollution? How many types of pollution? Define it

Give 10 examples for herbs , shrubs , climbers , creepers

Which state has the longest coastline in India A Tamil class 10 social science CBSE

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