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  • How to solve this problem- - is equal to : - Limit , continuity and differentiability - JEE Main

\lim_{x\rightarrow 1-}\frac{\sqrt{\pi }-\sqrt{2\sin ^{-1}x}}{\sqrt{1-x}} is equal to : 

  • Option 1)

     

    \sqrt{\frac{2}{\pi }}

     

     

     

  • Option 2)

     

    \sqrt{\pi }

  • Option 3)

     

    \sqrt{\frac{\pi }{2}}

  • Option 4)

     

    \frac{1}{\sqrt{2\pi }}

Answers (1)

best_answer

 

Limit of product / quotient -

Limit of product/quotient is the product/quotient of individual limits such that

\lim_{x\rightarrow a}{\left (f(x).g(x) \right )}

=\lim_{x\rightarrow a}{f(x).\lim_{x\rightarrow a}g(x), given that f(x) and g(x) are non-zero finite values

\lim_{x\rightarrow a}\frac{f(x)}{g(x)}=\frac{\lim_{x\rightarrow a}f(x)}{\lim_{x\rightarrow a}g(x)}, given that f(x) and g(x) are non-zero finite values

Also\:\lim_{x\rightarrow a}{kf(x)} 

=k\lim_{x\rightarrow a}{f(x)}

 

-

 

 

Method of factorisation -

 Indeterminate\:form\:of\:\frac{0}{0}\:and \:\frac{\infty }{\infty }

We remove the denominator factor which it makes zero.


\lim_{x\rightarrow 1}\:\frac{x^{2}-1}{x-1}=\lim_{x\rightarrow 1}=\frac{(x-1)(x+1)}{(x-1)}=1+1=2

 

- wherein

\frac{0}{finite}=0


\frac{finite}{0}=\infty

 

\lim_{x\rightarrow 1-}\frac{\sqrt{x}-\sqrt{2sin^{-1}x}}{\sqrt{1-x}}\\let \: \: sin\; x=t\\\\\lim_{t\rightarrow \frac{\pi }{2}}\frac{\sqrt{x}-\sqrt{2t}}{1-sin\; t}=lim_{t\rightarrow \frac{\pi }{2}}\: \frac{x-2t}{(\sqrt{x}+\sqrt{2t}\sqrt{1-sin\: t})}\\\\\\=lim_{t\rightarrow \frac{\pi }{2}}\: \frac{2(\frac{x}{2}-t)}{2\sqrt{x}\sqrt{1-cos\left ( \frac{x}{2}-t \right )}}\\\\\\=\lim_{h\rightarrow 0}\frac{h}{\sqrt{x}\sqrt{1-cosh}}\\\\\sqrt{\frac{2}{x}}


Option 1)

 

\sqrt{\frac{2}{\pi }}

 

 

 

Option 2)

 

\sqrt{\pi }

Option 3)

 

\sqrt{\frac{\pi }{2}}

Option 4)

 

\frac{1}{\sqrt{2\pi }}

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