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  • Solve! - Limit , continuity and differentiability - JEE Main-5

Let f(x)= [x^{2}]  then at x= 0 

  • Option 1)

    limit exist but not equaln to f(0)

  • Option 2)

    Limit doesn't exist

  • Option 3)

    f(x) is discontinous 

  • Option 4)

    f(x) is continous 

 

Answers (1)

best_answer

As we have learned

Condition for discontinuity -

1. \:L\neq R

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

limit of function at x = a does not exist.

2.\:L=R\neq V

limit exist but not equal to  x = a

-

 

 f(x)=[x^{2}]\Rightarrow f(0)=0

LHL =\lim_{x\rightarrow 0^{-}}[x^{2}]=0; RHL = \lim_{x\rightarrow 0^{+}}[x^{2}]=0

\therefore LHL=RHL=f(0)

f(x) is continous at x=0

 

 

 

 


Option 1)

limit exist but not equaln to f(0)

Option 2)

Limit doesn't exist

Option 3)

f(x) is discontinous 

Option 4)

f(x) is continous 

Posted by

Himanshu

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