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  • Help me answer: - Limit , continuity and differentiability - JEE Main-8

Let f(x)=x^{2}.  Then at x=0 ,f(x) is 

  • Option 1)

    continous and diffrentiable

  • Option 2)

    continous and non-diffrentiable

  • Option 3)

    neither continous nor diffrentiable

  • Option 4)

    not continous but diffrentiable

 

Answers (1)

best_answer

 As we have learned

Differentiability -

Let  f(x) be a real valued function defined on an open interval (a, b) and  x\epsilon (a, b).Then  the function  f(x) is said to be differentiable at   x_{\circ }   if

\lim_{h\rightarrow 0}\:\frac{f(x_{0}+h)-f(x_{0})}{(x_{0}+h)-x_{0}}


or\:\:\:\lim_{h\rightarrow 0}\:\frac{f(x)-f(x_{0})}{x-x_{0}}

-

 

 For continuity 

\lim_{x\rightarrow 0^{-}}x^{2}= \lim_{x\rightarrow 0^{+}}x^{2}= f(0)=0 

f(x) is continous 

For diffrentiability 

\because f(x) is continous , so now we will find 

\lim_{x\rightarrow 0}\frac{f(x)-f(0)}{x-0}=\lim_{x\rightarrow 0}\frac{x^{2}-0}{x}=0=finite

\therefore This limit exists , hence diffrentiable 

 

 

 

 

 


Option 1)

continous and diffrentiable

Option 2)

continous and non-diffrentiable

Option 3)

neither continous nor diffrentiable

Option 4)

not continous but diffrentiable

Posted by

Himanshu

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