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  • Help me answer: - Limit , continuity and differentiability - JEE Main-7

f(x)= \left \{ [x^{2}] ; x\neq 0 : 1; x=0\right.     then 

 

  • Option 1)

    f(x) is continous at x=0

  • Option 2)

    f(x) doesn't posses existy limit at x=0

  • Option 3)

    f(x) has removable discontinuity at x= 0

  • Option 4)

    f(x) is continous at x=1

 

Answers (1)

best_answer

As we have learned

Removal discontinuity -

A function  f is said to possess removable discontinuity if at x = a :  L=R\neq V

 

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

- wherein

 

 f(0^{+})=0 ; f(0^{-})=0 ; f(0)=1

\therefore limit exist but not equal to f(0) 

\therefore it has removable discontinuity at x=0

 

f(1^{-})=0;f(1^{+}), f(1)=1

LHL\neq RHL  so f (x) is discontinous at x=1

 

 

 

 


Option 1)

f(x) is continous at x=0

Option 2)

f(x) doesn't posses existy limit at x=0

Option 3)

f(x) has removable discontinuity at x= 0

Option 4)

f(x) is continous at x=1

Posted by

Himanshu

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