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  • I have a doubt, kindly clarify. - Limit , continuity and differentiability - JEE Main-6

f(x)= [x]\forall n\epsilon R, then f(x) is discontinous at x= 2 because

  • Option 1)

    L.H.L =R.H.L but diffrent from f(x)

  • Option 2)

    L.H.L are not defined

  • Option 3)

     R.H.L are not defined 

  • Option 4)

    L.H.L , R.H.L and f(x) are not equal all together

 

Answers (1)

best_answer

As we have learned

Rule for continuous -

A function is continuous at  x = a if and only if 

        L=R=V

L.H.L    R.H.L   value at  x = a.

- wherein

Where 

L=\lim_{x\rightarrow a^{-}}\:f(x)

R=\lim_{x\rightarrow a^{+}}\:f(x)

V_{I}=\lim_{x\rightarrow a}\:f(x)

 

 \lim_{x\rightarrow 2^{-}}f(x)=\lim_{x\rightarrow 2^{-}}[x]=1=LHL

\lim_{x\rightarrow 2^{+}}f(x)=\lim_{x\rightarrow 2^{+}}[x]=2=RHL

f(2)=[2]=2

\therefore LHL , RHL and f(2) are all not equal together so 

 

 

 

 


Option 1)

L.H.L =R.H.L but diffrent from f(x)

Option 2)

L.H.L are not defined

Option 3)

 R.H.L are not defined 

Option 4)

L.H.L , R.H.L and f(x) are not equal all together

Posted by

Himanshu

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