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  • Solve! - Limit , continuity and differentiability - JEE Main-7

f(x)=\left \{ e^{-1/|x|} ; x\neq 0 : 1 ; x= 0\right.   then 

  • Option 1)

    f(x ) is continous at x= 0

  • Option 2)

    f(x) has non - exsiting limit at x= 0  

  • Option 3)

    f(x) has LHL=RHL = f(0)

  • Option 4)

    f(x) has removable discontinuty at x=0

 

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

 

 

LHL= \lim_{x\rightarrow 0^{-}}e^{-1/|x|}= \lim_{x\rightarrow 0^{-}}e^{1/x}=0

 

RHL= \lim_{x\rightarrow 0^{+}}e^{-1/|x|}= \lim_{x\rightarrow 0^{+}}e^{1/x}=0

f(0)=1

Limit exists but  not equal to f(0) 

removable discontinuity at x= 0

 

 

 


Option 1)

f(x ) is continous at x= 0

Option 2)

f(x) has non - exsiting limit at x= 0  

Option 3)

f(x) has LHL=RHL = f(0)

Option 4)

f(x) has removable discontinuty at x=0

Posted by

Himanshu

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