Math, asked by sss549, 4 months ago

lin x tends to 0 (√1+x) -(√1-x) /(sin^(-1)x)​

Answers

Answered by senboni123456
1

Step-by-step explanation:

We have,

 \lim_{x \rarr0} \frac{ \sqrt{1 + x}  -  \sqrt{1 - x} }{ \sin^{ - 1} (x) }  \\

 = \lim_{x \rarr0} \frac{ \sqrt{1 + x}  -  \sqrt{1 - x} }{x}  \times  \frac{x}{ \sin^{ - 1} (x) }  \\

 = \lim_{x \rarr0} \frac{(1 + x - 1 + x)}{x( \sqrt{1 + x}   +  \sqrt{1 - x} )}  \times  \frac{1}{\lim_{x \rarr0} \frac{ \sin^{ - 1} (x) }{x} }  \\

 = \lim_{x \rarr0} \frac{2x}{x( \sqrt{1 + x}  +  \sqrt{1 - x}) } \times 1 \\

 = \lim_{x \rarr0} \frac{2}{ \sqrt{1  + x}  +  \sqrt{1 - x} }  \\

 =  \frac{2}{2}  \\

 = 1

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