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  • I need help with - Limit , continuity and differentiability - JEE Main-7

\lim_{x\rightarrow \pi /2}\frac{\sqrt{1+ \sin x}- \sqrt2}{\cos ^{2 }x} equals

  • Option 1)

    \frac{-1}{4\sqrt2}

  • Option 2)

    \frac{1}{4\sqrt2}

  • Option 3)

    \frac{1}{2\sqrt2}

  • Option 4)

    \frac{-1}{2\sqrt2}

 

Answers (1)

best_answer

As we have learned

Method of Rationalisation -

Rationalisation method is used when we have RADICAL SIGNS in an expression.(like  1/2,  1/3 etc) and there exists a negative sign between two terms of an algebraic expression.

- wherein

\lim_{x\rightarrow a}\:\frac{x-a}{\sqrt{x}-\sqrt{a}}


\therefore \:\frac{(x-a)(\sqrt{x}+\sqrt{a})}{(\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a})}


=\sqrt{x}+\sqrt{a}

=\sqrt{a}+\sqrt{a}

=2\sqrt{a}

 

 =\lim_{x\rightarrow \pi /2} \frac{\sqrt{1+ \sin x}-\sqrt{2}}{\cos^{x}}\times \frac{\sqrt{1+ \sin x}+\sqrt{2}}{\sqrt{1+ \sin x}+\sqrt{2}}

=\lim_{x\rightarrow \pi /2} \frac{\sin x-1}{\cos^{2}x(\sqrt{1+ \sin x}+\sqrt{2})}=\lim_{x\rightarrow \pi /2}\frac{-1}{(1+ \sin x)(\sqrt{1+ \sin x}+\sqrt{2})}

=\frac{-1}{2*2\sqrt2}= \frac{-1}{4\sqrt2}

 

 

 

 

 


Option 1)

\frac{-1}{4\sqrt2}

Option 2)

\frac{1}{4\sqrt2}

Option 3)

\frac{1}{2\sqrt2}

Option 4)

\frac{-1}{2\sqrt2}

Posted by

Himanshu

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