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  • I have a doubt, kindly clarify. - Limit , continuity and differentiability - JEE Main-5

If f(x)= \left \{ \right. \frac{x^{2}-4}{x-2} ; x\neq 2 : 1 ; x=2;

then f(x)  is discontinous  at x= 2 because

                       

  • Option 1)

    L.H.L and R.H.L are finitely diffrent 

  • Option 2)

    f(2) is not defined

  • Option 3)

    L.H.L =R.H.L but not same as f(2)

  • Option 4)

    L.H.L ,R.H.L and f(2) all are distinct

 

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)

 

 LHL=\lim_{x\rightarrow 2^{-}}\frac{x^{2}-4}{x-2}=\lim_{x\rightarrow 2^{-}}(x+2)=4

RHL=\lim_{x\rightarrow 2^{+}}\frac{x^{2}-4}{x-2}=\lim_{x\rightarrow 2^{+}}(x+2)=4

f(2)=1

\therefore LHL=RHL\neq f(2)

 

 

 

 

 


Option 1)

L.H.L and R.H.L are finitely diffrent 

Option 2)

f(2) is not defined

Option 3)

L.H.L =R.H.L but not same as f(2)

Option 4)

L.H.L ,R.H.L and f(2) all are distinct

Posted by

Himanshu

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