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Hi, Please answer with proper steps: Reply as soon as possible.

Hi,
Please answer with proper steps:
Reply as soon as possible.

Grade:11

1 Answers

Pintu Chaudhary
37 Points
2 years ago
We know that the function is continuous at x=c only when the functional value is equal to limiting value. i.e, 
                     
                                                                \lim_{x\rightarrow c}=f(c)
 
so,        (\lim_{x\rightarrow 0}) \frac{x}{1-\sqrt{1-x}}      = f(0)
   
and 
(\lim_{x\rightarrow 0}) \frac{x}{1-\sqrt{1-x}}   
   =   (\lim_{x\rightarrow 0}) \frac{x}{1-\sqrt{1-x}}  \tfrac{1+\sqrt{1-x}}{1+\sqrt{1-x}} 
  = (\lim_{x\rightarrow 0})\tfrac{x(1+\sqrt{1-x})}{1-1+x}
  =  (\lim_{x\rightarrow 0})1+\sqrt{1-x}
   =1+1=2
 
,therefore f(0)=2 Ans.
 

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