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  • Solve! - Limit , continuity and differentiability - JEE Main-6

Let f(x)  f(x)=\begin{Bmatrix} e^{1/x} & x \neq0\\ 0&x=0 \end{Bmatrix}  

then 

  • Option 1)

    f(x)  is continous at x= 0

  • Option 2)

    f(x)  is continous from left  at x= 0

  • Option 3)

    f(x)  is continous from right at x= 0

  • Option 4)

    Limit exists at x=0 but not equal to f(0)

 

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

-

 

 \lim_{x\rightarrow 0^{-}}f(x)= \lim_{x\rightarrow 0^{-}}e^{1/x}=0

\lim_{x\rightarrow 0^{+}}f(x)= \lim_{x\rightarrow 0^{+}}e^{1/x}=\infty

f(0)=0

For (A) All three are not equal to disconitnous at x=0 

For (B) LHL = f(0) \therefore it will be continous from left at x=0

For (C) LHL \neq f(0) \therefore it will be continous from right at x=0

For (D) LHL \neq RHL  so limit doesn't exist 

 

 

 

 

 

 

 

 


Option 1)

f(x)  is continous at x= 0

Option 2)

f(x)  is continous from left  at x= 0

Option 3)

f(x)  is continous from right at x= 0

Option 4)

Limit exists at x=0 but not equal to f(0)

Posted by

Himanshu

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